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r UNIVERSITY -OF CALIFORNIA AT LOS ANGELES INTKODUCTION TO QUATERNIONS, WITH NUMEROUS EXAMPLES. INTRODUCTION TO QUATEENIONS, WITH NUMEEOUS EXAMPLES. BY P. KELLAND, M.A, F.R.S, FORMERLY FELLOW OF QUEENS* COLLEGE, CAMBRIDGE ; AND P. G. TAIT, M.A., SEC. R.S.E., FORMERLY FELLOW OF ST PETER'S COLLEGE, CAMBRIDGE J PROFESSORS IN THE DEPARTMENT OF MATHEMATICS IN THE UNIVERSITY OF EDINBURGH. SECOND EDITION: ...,*; Honlron : MACMILLAN AND CO. 1882 [All Eights reserved.] FEINTED BY C. J. CLAY, M.A. AT THE UNIVERSITY PRESS. Mitkemstictl Sciences Library K28i PREFACE. Ix preparing this second edition for press I have altered as slightly as possible those portions of the work which were written entirely by Prof. Kelland. The mode of presentation which he employed must always be of great interest, if only from the fact that he was an exceptionally able teacher ; but the success of the work, as an introduction to a method which is now rapidly advancing in general estimation, would of itself have been a sufficient motive for my refraining from any serious alteration. A third reason, had such been necessary, would have pre- sented itself in the fact that I have never considered with the necessary care those metaphysical questions connected with the growth and development of mathematical ideas, to which my late venerated teacher paid such particular attention. My own part of the book (including mainly Chap. X. and worked out Examples 10 24 in Chap. IX.) was written hurriedly, and while I was deeply engaged with work of a very different kind ; so that I had no hesitation in determining to re-cast it where I fancied I could improve it. P. G. TAIT. UNIVERSITY OP EDINBUEOH, November, 1881. 210949 274 PEEFACE TO THE FIRST EDITION. THE present Treatise is, as the title-page indicates, the joint production of Prof. Tait and myself. The preface I write in the first person, as this enables me to offer some personal explanations. For many years past I have been accustomed, no doubt very imperfectly, to introduce to my class the subject of Quaternions as part of elementary Algebra, more with the view of establishing principles than of applying processes. Experience has taught me that to induce a student to think for himself there is nothing so effectual as to lay before him the different stages of the development of a science in some- thing like the historical order. And justice alike to the student and the subject forbade that I should stop short at that point where, more simply and more effectually than at any other, the intimate connexion between principles and pro- cesses is made manifest. Moreover, in lecturing on the ground- work on which the mathematical sciences are based, I could not but bring before my class the names of great men who spoke in other tongues and belonged to other nationalities than their own Diophantus, Des Cartes, Lagrange, for in- stance and it was not just to omit the name of one as Vlll PREFACE. great as any of them, Sir William Kowan Hamilton, who spoke their own tongue and claimed their own nationality. It is true the name of Hamilton has not had the impress of time to stamp it with the seal of immortality. And it must be admitted that a cautious policy which forbids to wander from the beaten paths, and encourages converse with the past rather than interference with the present, is the true policy of a teacher. But in the case before us, quite irrespective of the nationality of the inventor, there is ample ground for introducing this subject of Quaternions into an elementary course of mathematics. It belongs to first principles and is their crowning and completion. It brings those principles face to face with operations, and thus not only satisfies the student of the mutual dependence of the two, but tends to carry him back to a clear apprehension of what he had probably failed to appreciate in the sub- ordinate sciences. Besides, there is no branch of mathematics in which results of such wide variety are deduced by one uniform process; there is no territory like this to be attacked and subjugated by a single weapon. And what is of the utmost importance in an. educational point of view, the reader of this subject does not require to encumber his memory with a host of conclusions already arrived at in order to advance. Every problem is more or less self- contained. This is my apology for the present treatise. The work is, as I have said, the joint production of Prof. Tait and myself. The preface I have written without consulting my colleague, as I am thus enabled PREFACE. ix to say what could not otherwise have been said, that mathematicians owe a lasting debt of gratitude to Prof. Tait for the singleness of purpose and the self-denying zeal with which he has worked out the designs of his friend Sir Wm. Hamilton, preferring always the claims of the science and of its founder to the assertion of his own power and originality in its development. For rny own part I must confess that my knowledge of Quaternions is due exclusively to him. The first work of Sir Wm. Hamilton, Lectures on Quaternions, was very dimly and im- perfectly understood by me and I dare say by others, until Prof. Tait published his papers on the subject in the Messenger of Mathematics. Then, and not till then, did the science in all its simplicity develope itself to me. Sub- sequently Prof. Tait has published a work of great value and originality, An Elementary Treatise on Quaternions. The literature of the subject is completed in all but what relates to its physical applications, when I mention in addition Hamilton's second great work, Elements of Quater- nions, a posthumous work so far as publication is concerned, but one of which the sheets had been corrected by the author, and which bears all the impress of his genius. But it is far from elementary, whatever its title may seem to imply; nor is the work of Prof. Tait altogether free from difficulties. Hamilton and Tait write for mathematicians, and they do well, but the time has come when it behoves some one to write for those who desire to become mathe- maticians. Friends and pupils have urged me to undertake this duty, and after consultation with Prof. Tait, who from X PREFACE. being my pupil in youth is my teacher in riper years, I have, in conjunction with him, and drawing unreservedly from his writings, endeavoured in the first nine chapters of this treatise to illustrate and enforce the principles of this beautiful science. The last chapter, which may be regarded as an introduction to the application of Quater- nions to the region beyond that of pure geometry, is due to Prof. Tait alone. Sir W. Hamilton, on nearly the last completed page of his last work, indicated Prof. Tait as eminently fitted to carry on happily and usefully the appli- cations, mathematical and physical, of Quaternions, and as likely to become in the science one of the chief successors of its inventor. With how great justice, the reader of this chapter and of Prof. Tait's other writings on the subject will judge. PHILIP KELLAND. UNIVEESITT OF EDINBURGH, October, 1873. CONTENTS. CHAPTER L PAGES INTRODUCTORY 1 5 CHAPTER II. VECTOR ADDITION AND SUBTRACTION 6 31 Definition of a VECTOR, with conclusions immediately resulting therefrom, Art. 1 6 ; examples, 7 ; definition of UNIT VECTOR and TENSOR, with examples, 8; coplanarity of three coinitial vectors, with conditions requisite for their terminating in a straight line, and examples, 9 13 ; mean point, 14. ADDITIONAL EXAMPLES TO CHAPTER IL CHAPTER m. VECTOR MULTIPLICATION AND DIVISION 32 68 Definition of multiplication, and first principles, Art. 15 18 ; fundamental theorems of multiplication, 19 22 ; examples, 23; definitions of DIVISION, VERSOR and QUATERNION, 24 28 ; examples, 29 ; conjugate quaternions, 30 ; interpretation of formulae, 31. ADDITIONAL EXAMPLES TO CHAPTER IIL Xll CONTENTS. CHAPTER IV. PAGES THE STRAIGHT LINE AND PLANE 59 72 Equations of a straight line and plane, 32, 33 ; modifications and results length of perpendicular on a plane condition that four points shall lie in the same plane, &c. 34; examples, 35. ADDITIONAL EXAMPLES TO CHAPTER IV. CHAPTER V. THE CIRCLE AND SPHERE . . 7390 Equations of the circle, -with examples, 36, 37; tangent to circle and chord of contact, 38, 39; examples, 40 ; equations of the sphere, with examples, 41, 42. ADDITIONAL EXAMPLES TO CHAPTER V. CHAPTER VI. THE ELLIPSE . . . . . 91105 Equations of the ellipse, 43 ; properties of <pp, 44 ; equation of tangent, 45; Cartesian equations, 46; (j>~ l p, ^p, &c. 47; properties of the ellipse, -with examples, 48 50. ADDITIONAL EXAMPLES TO CHAPTER VI. CHAPTER VII. THE PARABOLA AND HYPERBOLA 106 127 -\ Equation of the parabola in terms of <j>p, with examples, 52 54; equations of the parabola, ellipse and hyperbola in a form corre- sponding to those with Cartesian co-ordinates, with examples, 55. ADDITIONAL EXAMPLES TO CHAPTER VII. CONTENTS. xiii CHAPTEE VIII. PAGES CENTRAL SURFACES OF THE SECOND ORDER 128 153 Equation of the ellipsoid, 56; tangent plane and perpendicular on it, 57, 58; polar plane, 59, 60; conjugate diameters and diame- tral planes, with examples, 60 64 ; the cone, 65, 66 ; examples on central surfaces, 67 ; Pascal's hexagram, 68 ADDITIONAL EXAMPLES TO CHAPTER VIII. CHAPTEE IX. FORMULAE AND THEIR APPLICATION 154 181 Formulas, 69, 70; examples, 71. ADDITIONAL EXAMPLES TO CHAPTER IX. CHAPTEE X. VECTOR EQUATIONS OF THE FIRST DEGREE 182 212 APPENDIX . 213232 INTRODUCTION TO QUATERNIONS. CHAPTER I. INTRODUCTORY. THE science named Quaternions by its illustrious founder, Sir William Rowan Hamilton, is the last and the most beautiful ex- ample of extension by the removal of limitations. The Algebraic sciences are based on ordinary arithmetic, start- ing at first with all its restrictions, but gradually freeing themselves from one and another, until the parent science scarce recognises itself in its offspring. A student will best get an idea of the thing by considering one case of extension within the science of Arith- metic itself. There are two distinct bases of operation in that science addition and multiplication. In the infancy of the science the latter was a mere repetition of the former. Multiplication was, in fact, an abbreviated form of equal additions. It is in this form that it occurs in the earliest writer on arithmetic whose works have come down to us Euclid. Within the limits to which his prin- ciples extended, the reasonings and conclusions of Euclid in his seventh and following Books are absolutely perfect. The demon- stration of the rule for finding the greatest common measure of two numbers in Prop. 2, Book VII. is identically the same as that which is given in all modern treatises. But Euclid dares not venture on fractions. Their properties were probably all but un- known to him. Accordingly we look in vain for any demonstration of the properties of fractions in the writings of the Greek arith- meticians. For that we must come lower down. On the revival T. Q. 1 2 QUATERNIONS. [CHAP. of science in the West, we are presented with categorical treatises on arithmetic. The first printed treatise is that of Lucas de Burgo in 1494. The author considers a fraction to be a quotient, and thus, as>he expressly states, the order of operations becomes the reverse of that for whole numbers multiplication precedes addi- tion, etc. In our own country we have a tolerably eai'ly writer on arithmetic, Robert Record, who dedicated his work to King Edward the Sixth. The ingenious author exhibits his treatise in the form of a dialogue between master and scholar. The scholar battles long with this difficulty that multiplying a thing should make it less. At first, the master attempts to explain the anomaly by reference to proportion, thus : that the product by a fraction bears the same proportion to the thing multiplied that the multiplying fraction does to unity. The scholar is not satisfied ; and accord- ingly the master goes on to say : "If I multiply by more than one, the thing is increased ; if I take it but once, it is not changed; and if I take it less than once, it cannot be so much as it was before. Then, seeing that a fraction is less than one, if I multiply by a fraction, it follows that I do take it less than once", etc. The scholar thereupon replies, " Sir, I do thank yon much for this reason ; and I trust that I do perceive the thing". Need we add that the same difficulty which the scholar in the time of King Edward experienced, is experienced by every thinking boy of our own times; and the explanation afforded him is precisely the same admixture of multiplication, proportion, and division which suggested itself to old Robert Record. Every schoolboy feels that to multiply by a fraction is not to multiply at all in the sense in which multiplication was originally presented to him, viz. as an abbreviation of equal additions, or of repetitions of the thing multi- plied. A totally new view of the process of multiplication has insensibly crept in by the advance from whole numbers to fractions. So new, so different is it, that we are satisfied Euclid in his logical and unbending march could never have attained to it. It is only by standing loose for a time to logical accuracy that extensions in the abstract sciences extensions at any rate which stretch from one science to another are effected. Thus Diophantus in his I.] INTRODUCTORY. 3 Treatise on Arithmetic (i.e. Arithmetic extended to Algebra) boldly lays it down as a definition or first principle of his science that 'minus into minus makes plus'. The science he is founding * O is subject to this condition, and the results must be interpreted consistently with it. So far as this condition does not belong to ordinary arithmetic, so far the science extends beyond ordinary arithmetic : and this is the distance to which it extends It makes subtraction to staud by itself, apart from addition; or, at any rate, not dependent on it. "We trust, then, it begins to be seen that sciences are extended by the removal of barriers, of limitations, of conditions, on which sometimes their very existence appears to depend. Fractional arithmetic was an impossibility so long as multiplication was re- garded as abbreviated addition ; the moment an extended idea was entertained, ever so illogically, that moment fractional arithmetic started into existence. Algebra, except as mere symbolized arith- metic, was an impossibility so long as the thought of subtraction was chained to the requirement of something adequate to subtract from. The moment Diophantus gave it a separate existence boldly and logically as it happened by exhibiting the law of minus in the forefront as the primary definition of his science, that moment algebra in its highest form became a possibility ; and indeed the foundation-stone was no sooner laid than a goodly building arose on it. The examples we have given, perhaps from their very simplicity, escape notice, but they are not less really examples of extension from science to science by the removal of a restriction. We have selected them in preference to the more familiar one of the extension of the meaning of an index, whereby it becomes a logarithm, because they prepare the way for a further extension in the same direction to which we are presently to advance. Observe, then, that in frac- tions and in the rule of signs, addition (or subtraction) is very slenderly connected with multiplication (or division). Arithmetic as Euclid left it stands on one support, addition only, inasmuch as with him multiplication is but abbreviated addition. Arithmetic in its extended form rests on two supports, addition and multiplica- 12 4 QUATERNIONS. [CHAP. tion, the one different from the other. This is the first idea we want our reader to get a firm hold of ; that multiplication is not necessarily addition, but an operation self-contained, self-interpret- able springing originally out of addition ; but, when full-grown, existing apart from its parent. The second idea we want our reader to fix his mind on is this, that when a science has been extended into a new form, certain limitations, which appeared to be of the nature of essential truths in the old science, are found to be utterly untenable ; that it is, in fact, by throwing these limitations aside that room is made for the growth of the new science. We have instanced Algebra as a growth out of Arithmetic by the removal of the restriction that subtraction shall require something to subtract from. The word 'subtraction' may indeed be inappropriate, as the word multiplication ap- peared to be to Record's scholar, who failed to see how the multi- plication of a thing could make it less. In the advance of the sciences the old terminology often becomes inappropriate ; but if the mind can extract the right idea from the sound or sight of a word, it is the part of wisdom to retain it. And so all the old words have been retained in the science of Quaternions to which we are now to advance. The fundamental idea on which the science is based is that of motion of transference. Real motion is indeed not needed, any more than real superposition is needed in Euclid's Geometry. An appeal is made to mental ti'ansference in the one science, to mental superposition in the other. We are then to consider how it is possible to frame a new science which shall spring out of Arithmetic, Algebra, and 'Geometry, and shall add to them, the idea of motion of transference. It must be confessed the project we entertain is not a project due to the nineteenth century. The Geometry of Des Cartes was based on something very much resembling the idea of motion, and so far the mere introduction of the idea of transference was not of much value. The real advance was due to the thought of severing multiplication from addition, so that the one might be the representative of a kind of motion absolutely different from that which was represented by I.] INTRODUCTORY. 5 the other, yet capable of being combined with it. What the nine- teenth century has done, then, is to divorce addition from multipli- cation in the new form in which the two are presented, and to cause the one, in this new character, to signify motion forwards and backwards, the other motion round and round. We do not purpose to give a history of the science, and shall accordingly content ourselves with saying, that the notion of sepa- rating addition from multiplication attributing to the one, motion from a point, to the other motion about a point had been floating iu the minds of mathematicians for half a century, without producing many results worth recording, when the subject fell into the hands of a giant, Sir William Rowan Hamilton, who early found that his road was obstructed- he knew not by what obstacle so that many points which seemed within his reach were really inaccessible. He had done a considerable amount of good work, obstructed as he was, when, about the year 1843, he perceived clearly the obstruction to his progress in the shape of an old law which, prior to that time, had appeared like a law of common sense. The law in question is known as the commutative law of multiplication. Presented in its simplest form it is nothing more than this, ' five times three is the same as three times five'; more generally, it appears under the form of 'ab = ba whatever a and b may represent'. When it came distinctly into the mind of Hamilton that this law is not a necessity, with the extended signification of multiplication, he saw his way clear, and gave up the law. The barrier being removed, he entered on the new science as a warrior enters a besieged city through a practicable breach. The reader will find it easy to enter after him. CHAPTER II. VECTOR ADDITION AND SUBTRACTION. 1. Definition of a Vector. A vector is the representative of transference through a given distance, in a given direction. Thus if AB be a straight line, the idea to be attached to 'vector AB' is that of transference from A to B. For the sake of definiteness we shall frequently abbreviate the phrase ' vector AB ' by a Greek letter, retaining in the meantime (with one exception to be noted in the next chapter) the English letters to denote ordinary numerical quantities. If we now start from .Band advance to (7 in the same direction, BC being equal to AB, we may, as in ordinary geometry, designate ' vector EG ' by the same symbol, which we adopted to designate ' vector AB.' Further, if we start from any other point in space, and advance from that point by the distance OX equal to and in the same direction as AB, we are at liberty to designate 'vector OX' by the same symbol as that which represents AB. Other circumstances will determine the starting point, and in- dividualize the line to which a specific vector corresponds. Our definition is therefore subject to the following condition : All lines which are equal and drawn in the same direction are represented by the same vector symbol. We have purposely employed the phrase 'drawn in the same direction ' instead of ' parallel,' because we wish to guard the student against confounding 'vector AB ' with 'vector BA.' ART. 2.] VECTOR ADDITION AND SUBTRACTION. 7 2. In order to apply algebra to geometry, it is necessary to impose on geometry the condition that when a line measured in one direction is represented by a positive symbol, the same line measured in the opposite direction must be represented by the cor- responding negative symbol. In the science before us the same condition is equally requisite, and indeed the reason for it is even more manifest. For if a transference from A to B be represented by + a, the transference which neutralizes this, and brings us back again to A, cannot be conceived to be represented by anything but a, provided the symbols + and are to retain any of their old algebraic meaning. The vector AB, then, being represented by + a, the vector BA will be represented by - a. 3. Further it is abundantly evident that so far as addition and subtraction of parallel vectors are concerned, all the laws of Algebra must be applicable. Thus (in Art. 1) AB + BC or a + a produces the same result as AC which is twice as great as AB, and is there- fore properly represented by 2a ; and so on for all the rest. The distributive law of addition may then be assumed to hold in all its integrity so long at least as we deal with vectors which are paralk-1 to one another. In fact there is no reason whatever, so far, why a should not be treated in every respect as if it were an ordinary algebraic quantity. It need scarcely be added that vectors in the same direction have the same proportion as the lines which corre- spond to them. We have then advanced to the following LEMMA. All lines drawn in the same direction are, as vectors, to be represented by numerical multiples of one and the same symbol, to which the ordinary laws of Algebra, so far as their addi- tion, subtraction, and numerical multiplication are concerned, may be unreservedly applied. 4. The converse is of course true, that if lines as vectors are represented by multiples of the same vector symbol, they are parallel. 8 QUATERNIONS. [CHAP. II. It is only necessary to add to what has preceded, that if BC be a line not in the same direction with c AB, then the vector EG cannot be represented by a or by any (arith- metical) multiple of a. The vector A symbol a must be limited to express transference in a certain direction, and cannot, at the same time, express transference in any other direction. To express ' vector BC 1 then, another and quite independent symbol (3 must be introduced. This symbol, being united to a by the signs + and , the laws of algebra will, of course, apply to the combination. 5. If we now join AC, and thus form a triangle ABC, and if we denote vector AB by a, BC by ft, AC by y, it is clear that we shall be presented with the equation a + ft = y. This equation appears at first sight to be a violation of Euclid I. 20 : " Any two sides of a triangle are together greater than the third side". But it is not really so. The anomalous appearance arises from the fact that whilst we have extended the meaning of the symbol + beyond its arithmetical signification, we have said nothing about that of a symbol = . It is clearly necessary that the signification of this symbol shall be extended along with that of the other. It must now be held to designate, as it does perpetually in algebra, ' equivalent to.' This being premised, the equation above is f ree.d from its anomalous appearance, and is perfectly con- sistent with everything in ordinary geometry. Expressed in words it reads thus : ' A transference from A to B followed by a ti-ans- ference from B to C is equivalent to a transference from A to C.' 6. AXIOM. If two vectors have not the same direction, it is impossible that the one can neutralize the other. This is quite obvious, for when a transference has been effected from A to B, it is impossible to conceive that any amount of trans- ference whatever along BC can bring the moving point back to A. It follows as a consequence of this axiom, that if a, (3 be different actual vectors, i. e. finite vectors not in the same direction, and if ART. 7.] "VECTOR ADDITION AND SUBTRACTION. 9 ma. + n{3 = 0, where m and n are numerical quantities ; then must m and n = 0. Another form of this consequence may be thus stated. If [still with the above assumption as to a and /?] ma + n/3 = pa + q(3, then must mp, and n q. 7. We now proceed to exemplify the principles so far as they have hitherto been laid down. It is scarcely necessary to remind the reader that we are assuming the applicability of all the rules of algebra and arithmetic, so far as we are yet in a position to draw on them ; and consequently that our demonstrations of certain of Euclid's elementary propositions must be accepted subject to this assumption. To avoid prolixity, we shall very frequently drop the word vector, at least in cases where, either from the introduction of a Greek letter as its representative, or from obvious considerations, it must be clear that the mere line is not meant. The reader will not fail to notice that the method of demonstration consists mainly in reach- ing the same point by two different routes. (See remark on Ex. 9.) EXAMPLES. Ex. 1. Tlie, straight lines which join the extremities of equal and parallel straight lines towards the same parts are themselves equal and parallel. Let A E be equal and parallel to CD ; to prove that AC is equal and parallel to ED. Let vector AB be represented by a, then (Art. 1) vector CD is also, repre- sented by a. If now vector CA be represented by (3, vector DB by y, we shall have (Art. 5) vector CB = CA + A B = /3 + a, and vector CB = CD + DB = a + y ; . '. ft + a a + y, and (3 = y ; so that (3 and y are the same vector symbol; consequently (Art. 1) 10 QUATERNIONS. [CHAP. TI. the lines which they represent are equal and parallel ; i. e. CA is equal and parallel to ED. Ex. 2. The opposite sides of a parallelogram are equal; and the diagonals bisect each other. Since AB is parallel to CD, if vector AB be represented by a, vector CD will be represented by some numerical multiple of a (Art. 3), call it ma.. And since CA is parallel to DB; if vector CA be /3, then vector DB is nfi ; hence vector CB = CA+AB = p + a., and = CD + DB = ma + nfi ; .-. a + ft = ma + n{3. Hence (Art. G) m= 1, n=l, i.e. the opposite sides of the paral- lelogram are equal. Again, as vectors, AO + OB= AB = CD = CO + OD ; And as AO is a vector along OD, and CO a vector along OB ; it follows (Art. 6) that vector AO is vector OD, and vector CO is OB; O = OD CO = OB. Ex. 3. The sides about the. equal angles of equiangular triangles are proportionals^ Let the triangles ABC, A DE have a common angle A, then, because the angles D and B are equal, DE is parallel to BC. Let vector AD be represented by a, DE by /?, then (Art. 3) AB is ma, BC n/3. . -. as vectors, AE = AD + DE = a + (3, D B Now AC is a multiple of AE, call it p(a+{3).- . : ma + n[$ = p (a. + (3), and m p n (Art. 6). EX. 4.] VECTOR ADDITION" AND SUBTRACTION. 11 But line A B : AD = m, line EC : Z>E = n, .-. AB : AD :: EC : DE. Ex. 4. The bisectors of the sides of a triangle meet in a point which trisects each of them. Let the sides of the triangle ABC be bisected in D, E, F ; and let AD, BE meet in G. Let vector ED or DC be a, CE or EA (3, F ' then, as vectors, BA = EC + CA = 2a + 2/? = 2 (a + 0), hence (Art. 4) .5^1 is parallel to Z)^, and equal to '2DE. Again, G+GA= BA Now vector _36r is along GE, and vector 6-M along DG. .-. (Art. 6) .# GA = whence the same is true of the lines. 2 Lastly, BG = ^BE 12 QUATERNIONS. [CHAP. II. GF=BF-BG lience CG is in the same straight line with GF, and equal to IGF. Ex. 5. When, instead of D and E being the middle points of the sides, they are any points whatever in those sides, it is required to find G and the point in which CG produced meets AB. BG CA Let nr , = m, rr^ n \ a ^ so let vector Z>(7 = a, vector CE ~ (3 ; JL/O L/Jli .-. BG = ma, CA = n[J. Hence BE= BG + CE = Let BG = xBE, GA=yDA, then BA=BG + GA = x (ma + ft) + y (a -f n/3). But BA =ma + n(3, . '. (Art. G) xm + y = . BG (m-l)n AG (n-l)m QYirl 'VIA 11 C\\* - * 1 I I lAsm 1. *5* _ _-_ U \J1- . T , , BE mn-l AD mn-\. Again, let BF=pBA p (ma + nft). T) i T> Jjl ~)f*1 i /^ 7^ JjUC JjJ. 1 = .DO + O-r = ma + a multiple of CG = ma + zCG suppose = ma + z ' -=- (ma + 8) - 1 ( mn I The two values of BF being equated, and Art. 6 applied, there results ??, - 1 m 1 ~ 7/w 1 ' w 1 ' EX. 6.] VECTOR ADDITION AND SUBTRACTION. 13 whence i.e. I p n 1 p nt- 1 ' AF AE BD or AF . BD.CE=AE.CD. BF. Ex. 6. When, instead of as in Ex. 4, where D, E, F are points taken within BC, CA, AB at distances equal to half those lines respectively, they are points taken in BC, CA, AB produced, at the same distances respectively from C, A, and B ; to find tJie inter- sections. Let the points of intersection be respectively G iy G a , G 3 . E F^"G' Retaining the notation of Ex. 4, we have = 3a, CE=3/3; and .-. and = 3a + yD A . . 2x = 3 y, 3x = 2y, and x = = : 7 .-. line EG 3 =]=EB. 14 QUATERNIONS. [CHAP. II. Similarly line FG l = l - FC, \m& DG t =]=DA, f and from equation (1) EG & = (2a + 3/3). But BG a = BA + AG 3 = 2a + 2ft + AG, ; 2 hence line J 6r = line DA and similarly of the others. Ex. 7. :77ie middle points of the lines which join the points of bisection of the opposite sides of a quadrilateral coincide, whether the four sides of the quadrilateral be in the same plane or not. Let A BCD be a quadrilateral ; E, II, G, F the middle points of AB, EG, CD, DA X the middle point of EG. Let vector AB a, AC = ft, AD = y, then AE + G = AD + DG gives / = i (* + ft + y), which being symmetrical is a, ft, y in the same as the vector to the middle point of HF. X is called (Art. 14) the mean point of ABCD. Ex. 8. The point of bisection of the line which joins the middle points of the diagonals of a quadrilateral (plane or not) is the mean point. EX. 9.] VECTOR ADDITION AND SUBTRACTION. Let P, Q be the middle points of AC, BD, R that of PQ. Retaining the notation of the last ex- ample we have AR=(AP + AQ) 2i Similarly i.e. is the same point as X in the last example ; and is therefore the mean point of A BCD. Ex. 9. AD is drawn bisecting BC in D and is produced to any point E ; AB, CE prodded meet in P ; AC, BE in Q ; PQ is parallel to BC. Let AB = a, AC = ft, and AE is a multiple of AD ~ z (a + ft) say. Then CP =pC gives xa - ft =p [z (a + j3) - ft], .'. (Art. 6) x pz, 1 =pz p ; .. p = x+ 1. Similarly BQ - qBE gives y(3 a = q {z (a + (3) - a}, y = qz, -\=qz-q, 16 QUATERNIONS. [CHAP. ii. i x V i and since z = - = we have p y hence the line PQ is parallel to BG. The method pursued in this example leads to the solution of all similar problems. It consists, as we have already stated, in reach- ing the points P and Q respectively by two different routes, viz. through C and through E i or P ; through B and through E for Q and comparing the results. Cor. 1. PE : EC :: p-\ : 1 :: x : 1 :: AP : AB. Cor. 2. AE : AD :: 2z : .1 :: 2x : x+ I :: 2(p-l) :p :: 1PE : PC, .-. AD : DE :: PE + EC : PE-EG. Ex. 10. If DEF be drawn cutting the sides of a triangle ; then will AD.BF.CE = AE.CF. BD. Let BD = a, DA =pa, AE= ft, EG = and CF is a multiple of BG. Let CF= xBO CF=CE + EF =-EC+EF But .'. equating, we have x (1 + p} = yp, x(\+q whence x ( 1 +x)pq, CF BF AD CE 1 a _ _ _ BG~ BG' BD' AE' .-. AD.BF.CE = AE.CF.BD. EX. 11.] VECTOR ADDITION AND SUBTRACTION. 17 Ex. 11. If from any point within a parallelogram, parallels be drawn to the sides, the corresponding diagonals of the two parallelograms thus formed, and of the original parallelogram shall meet in the same point. Let PQ, US meet in T; join TO, OD. Let OA = a, OB = p, OQ=ma, OS=np, and also FO = TS equating, there results = TQ-OQ = x{np+(l-m)a}-ma, mm- ym ; and mn m 1 m w' mn . , l-m-n^ ^' I-m-n hence (Art. 4) TO, OD are in the same straight line. COR. TO : TD :: mn : (l-m)(l-n) :: OSCQ : CRDP. Ex. 12. T7te points of bisection of the three diagonals of a com- plete quadrilateral are in a straight line. 1. Q. 2 18 QUATERNIONS. P, Q, R, the middle points of the diagonals of the complete quadrila- teral A BCD, are in a straight line. Let A = a,AD = (3, AE = ma, AF=n'{3; D = - ma. and gives whence and x (n(3 a) + y (ft ma) = (3 d, l, x + my=\, ml .'. X = mn 1 ' AP =\^-\{ 1 m (n 1) a + n (m 1) /^ 2 m/i- 1 .:AQ-AP = AR-AP= ^ (ma + nft), 1 2(wm-l) 9 limn _L-1 \ I" ' V < I llvlb -" J. I or vector PR is a multiple of vector PQ, and therefore they are in the same straight line. COR. Line PQ : PR' :: I : mn :: AE.AD : AE . AF :: triangle AED : triangle AEF. We shall presently exemplify a very elegant method due to Sir W. Hamilton of proving three points to be in the same straight line. ART. 8.] VECTOR ADDITION AND SUBTRACTION. 19 8. It is often convenient to take a vector of the length of the unit, and to express the vector under consideration as a numerical multiple of this unit. Of course it is not necessary that the unit should have any specified value ; all that is required is that when once assumed for any given problem, it must remain unchanged throughout the discussion of that problem. If the line AB be supposed to be a units in length, and the unit vector along AB be designated by a, then will vector AB be a (Art. 3). Sir William Hamilton has termed the length of the line in such cases, the TENSOR of the vector; so that the vector AB is the product of the tensor AB and the unit vector along AB. Thus if, as in the examples worked under the last article, we designate the vector AB by a, we may write a = TaUa, where To. is an abbre- viation for ' Tensor of the vector a ; Ua. for ' unit vector along a'. EXAMPLES. Ex. 1. If tJie vertical angle of a triangle be bisected by a straight line which also cuts the base, the segments at the base shall have the same ratio that the other sides of the triangle have to one another, Take unit vectors along AB, AC, which call a, /3 respectively : construct a rhombus p. APQR on them and draw its diagonal AR. Then since the diagonals of a rhombus bi- _ sect its angles, it is clear that the vector AD which bisects the angle A is a multiple of AR the diagonal vector of the rhombus. Now AR Now vector AB = ca, AC = bfi; using c, b as in ordinary geometry for the lengths of AB, AC. Hence BD = AD - AB = x (a + /?) - ca, and BD = yBC=y(AC-AB] - ca). 22 20 QUATERNIONS. [CHAP. II. Equating, x-c = -yc, x = yb; and BD : DC :: y : l-y :: c : b :: BA : AC. COR. If a, ft are unit vectors from A, and if 8 be another vector from A such that 8 = x(a + ft); then 8 bisects the angle between a and (3. Ex. 2. The three bisectors of the angles of a triangle meet in a point. Let AD, BE bisect A, B and meet in G ; CG bisects C. Let units along AH, AC, BC be a, ft, y, then as in the last example, AG~x(a + ft), BG = y(-a. + y). But ay = bft- ca, bft-ca and CG =AG-AC = x( a + ft)-bft, also CG^BG-BC bft - cd =/ a+ -ca\ / a be whence x = - - - , a + b + c and CG hence CG bisects the angle C (Cor. Ex. 1). AKT. 9.] VECTOR ADDITION AND SUBTRACTION. 21 9. If a, (3, y are non-parallel vectors in the same plane, it is always possible to find numerical values of a, b, c so that aa + b(3 + cy shall = 0. For a triangle can be constructed whose sides shall be parallel respectively to a, (3, y. Now if the vectors corresponding to those sides taken in order be aa, b{3, cy respectively, we shall have, by going round the triangle, 10. If a, fi, y are three vectors neither parallel nor in the same plane, it is impossible to find numerical values of a, b, c, not equal to zero, which shall render aa + bft + cy=Q. For (Art. 5) aa + b(3 can be represented by a third vector in the plane which contains two lines parallel respectively to a, /?. Now cy is not in that plane, therefore (Art. 6) their sum cannot equal 0. It follows that if aa + 6/3 + cy = and a, /?, y are not parallel vectors, they are in the same plane. 11. There is but one way of making the sum of multiples of a, (3, y (as in Art. 9) equal to 0. Let aa+b/3 + cy = 0, and also a + (3 + r = 0. By eliminating y we get (ar cp)a + (br - cq) ft = ; .% (Art. 6) ar cp, br = cg, or a : b : c :: p : q : r, so that the second equation is simply a multiple of the first. 12. If a, ft, y are coinitial, coplanar vectors terminating in a straight line, then the same values of a, b, c which render aa + 6/3 + cy - will also render a -t- b + c = 0. 22 QUATERNIONS. [CHAP. ii. Let vector OA = a, OB = ft, OC = y, ABC being a straight line ; then But AC is a multiple of AB, or y-a=p(p-a), i.e. (p l)a pft + y = 0. But (;?-l)-p+l=0; and as p 1, p, +1 correspond to a, b, c and satisfy the con- dition required, the proposition is proved generally (Art. 11). 13. Conversely, if a, fi, y are coinitial coplanar vectors, and if both aa + b(3 + cy = Q and a + b + c - 0, then do a, ft, y terminate in a straight line. For ay + by + cy = ; therefore by subtraction i.e. y a is a multiple of y y8, and therefore (Art. 4) in the same straight line with it: i.e. AC is in the same straight line with BC. (See Tait's Quaternions, 30.) EXAMPLES. Ex. 1. If two triangles are so situated that the lines which join corresponding angles meet in a point, then pairs of correspond- ing sides being produced will meet in a straight line. ABC, A'B'C' are the triangles; the point in which A' A, B'B, C'C meet; P, Q, R the points in which BC, BC', <fec. meet: PQR is a straight line. Let OA = a, OB = (3, OC = y, and BA = a - ft, BR = x(a.-p); 'A' = ma nft, B'R = y (ma nfi). EX. 2.] VECTOR ADDITION AND SUBTRACTION. 23 Now BB' = BR- B'R gives (n~\.}B = x(a f$) y (ma nf3) ; /. n 1 = x + ny, = x my, and x = : whence OR= OB + BR = B-~ (a- B) m n ^ _ n (m 1) /3 m (w 1) a m . Similarly, OP^^^lb^ ^ _m(p-l)a-p(m-l)y p m . : (m - n )(p-l)OR+ (v -p)(m- 1) OP + (p-m)(n-l)OQ = Q. And also (m -n)(p-l) + (n -p) (m - 1) + (p - m) (n-l) = 0, whence (Art. 13) P, Q, R are in the same straight line. Ex. 2. If a quadrilateral be divided into two quadrilaterals by any cutting line, the centres of the three shall lie in a straight line. Let Pj(?,$ 3 P 3 be the quadrilateral divided into two by the line P S Q 2 . Let the diagonals of P g Q a Q 3 P 3 fflteet in R^ and so of the others : R lt R g , R 3 are the centres. 24 QUATERNIONS. [CHAP. II. Produce P 3 P,, Q a Q 1 to meet in 0. Let unit vectors along OP, 0$ be denoted by a, ft ; and put OP, = w^o, OP, = m s a, OP a = m t a ; then OR 3 = OP 1 + P,^ 3 = 7rc,a + x (njl - w^a), and OR 3 =OQ^ + Q l R a = Equating, we have m^-m^x m^/, and and Q ^ = OT,W, (n, - n,) a + ra,, (TO, - m,ri, - m t w a Similarly, (m, - OR 3 + (mji, - m a n^ m l n l OR^ + (m 3 n 3 w^) m/i a OR S = 0. And also (mfo - m s n s ) m a n a + (m a n a - m 3 n 3 ) m^ + (m 3 n a -m 1 n l )m a n a = 0, whence (Art. 13) R I} R 2 , R a are in the same straight line. COR. R t , R 3 , R s will pass through provided the coefficients of a and /3 in the three vectors have the same proportion, i.e. provided I ___ !_ I ___ |_..!_JL. .!_! m t m a ' m a m a " w, n a ' n a n a ' Ex. 3. If AD, BE, CF be drawn cutting one another at any point G within a triangle, tJien FD, DE, EF shall meet the third sides of the triangle produced in points which lie in a straight line. Also the produced sides of the triangle s/utll be cut harmo- nically. EX. 3.] VECTOB ADDITION AND SUBTRACTION. If, as in Ex. 5, Art. 7, we put 25 we get, as in that example, AF : BF :: n-l : m-l; . .. BF= m ~ l . (ma + ), m+n- 2 ^ and FD=BD- BF= DM xFD, compared with erives - 2) a - (m-l)(n-2) (m-l)n rg 3 ' V i - J jg _} L___ n I m + n 2 m+n 2 and n-2 ft- n-l Again, FE=FA+AE = - {ma - (m - 2) $} m + n-2 ( 26 QUATERNIONS. [CHAP. II. And EL = xFE, compared with m gives y m(m-l) a. Thirdly, i)jV= xZ>^ = a; (a + /3), compared with = BN- BD = y (ma + nj3) -(m-l) a, m-I gives y m n and EN = - (ma + n(3). m-n^ Now (m -l)(n- 2) BM + (m - n) BN Also (m-l)(n- 2) + (m- n)- (m- 2) (n-I) = Q therefore BM, BN, BL are in a straight line (Art. 13). Further, CL = ^ CD, m 2 m-2 .-. CL : CD :: BL : BD, and BL is cut harmonically. Ex. 4. The point of intersection of bisectors of tJie sides of a triangle from the opposite angles, the point of intersection of per- pendiculars on tJie sides from the opposite angles, and the point of intersection of perpendiculars on tJie sides from their middle points, lie in a straight line which is trisected by the first of these points. 1. Let unit vector CJ3 = a, unit vector CA = (3, then, Ex. 4, Art. 7, CG = \ (aa + bft). ART. 14.] VECTOR ADDITION AND SUBTRACTION. 27 2. Let AH, BK perpendiculars on the A sides intersect in 0, /\\ ft then HA = bft-bacosC, / JAY = b(ft-a cos C), Now CO = CA +AO, and also = CB + SO, gives 6)3 + yb (ft - aa cos C) = aa + xa(a ft cos (7), 6 cos C a and CO = -. ~ {(6 a cos (7) a + (a - b cos C) ft}. sin 2 (7 lv 3. Let perpendiculars from I) and E (-Ex. 4, Art. 7) meet in X, then DX is a multiple of HA. . : CX= CD + DX = CE + EX gives ^ aa + v (ft - a cos C) = ^ bft + z (a. - ft cos C), 2t a b a cos C 2 sin 2 C (a b cos C) a + (b a cos (7) ft 2 sin 2 C ~~ 1 and C X = and also 2 + 1-3 = 0, .. X, 0, G are in a straight line. Also CO-CG=2 (CG - CX), or vector GO = 2 vector XG, and G trisects XO. 14. The vector to the mean point of any polygon is the mean of the vectors to the angles of the polygon. 28 QUATERNIONS. [CHAP. II. 1. Let be any point ; then in the figure of Ex. 4, Art. 7, we have, calling OA, a, OB, (3 and 00, y, OG=a+AG=ft+G=y+CG 1 9 because AG + BG + CG =-- 1 (AD + BE + CF) o = I {(AB + AC) + (BA + C) + (CA + CB)} = 0. 2. If OA, OB, OC, OD be a, ft, y, 8, in the figure of Ex. 7, Art. 7, we have = OH + HX=OH+l (OF- OH) 3. In the more general case we may define the mean point in a manner analogous to that adopted in mechanics to define the centre of inertia of equal masses placed at the angular points of the figure. Thus, if we take any rectangular axes OX, OY, and designate by a, ft unit vectors parallel to these axes; and by p,, p 4 , &c. the vectors to the different points; and if we write x^ y,; x ii y> & c - f r the Cartesian co-ordinates of the different points referred to those axes ; and define the mean point as the centre of inertia of equal masses placed at the angular points; the Cartesian co-ordinates of that point will be a?, + ,+ ... _ ~ ~ and its vector p = xa + yft. EX. 1.] VECTOR ADDITION AND SUBTRACTION. 29 Now p l = xp + yfi, p a = x,a + y, &c. g ^ , , ... m " = /> COE. 1. ( Pl - P ) + (p 2 -p) + (p 3 i. e. the sum of the vectors of all the points, drawn from the mean point, = 0. The extension of the same theorem to three dimensions is obvious. COR. 2. If we have another system of n points whose vectors are cr l , o- , &c. then the vector to the mean point is n If now T be the mean point of the whole system, we have T== Pi + P + +<r 1 + <r,+ ... or (m + n) r mp rw = 0, hence (13) T, p, cr terminate in a right line; or the general mean point is situated on the right line which connects the two partial mean points. ADDITIONAL EXAMPLES TO CHAP. II. 1. If P, Q, B, S be points taken in the sides AB, EG, CD, DA of a parallelogram, so that AP : AB :: BQ : BC, &c., PQRS will form a parallelogram. 2. If the points be taken so that AP = CR, BQ = DS, the same is true. 3. The mean point of PQRS is in both cases the same as that of ABCD. 30 QUATERNIONS. [CHAP. II. 4. If FQ'R'S' be another parallelogram described as in Ex. 1, the intersections of PQ, P'Q', <fec. shall be in the angular points of a parallelogram EFGH constructed from PQRS as P'Q'R'S' is constructed from ABGD. 5. The quadrilateral formed by bisecting the sides of a quadrilateral and joining the successive points of bisection is a parallelogram, with the same mean point. 6. If the same be true of any other equable division such as trisection, the original quadrilateral is a parallelogram. 7. If any line pass throvigh the mean point of a number of points, the sum of the perpendiculars on this line from the different points, measured in the same direction, is zero. 8. From a point E in the common base AB of the two triangles ABC, ABD, straight lines are drawn parallel to AC, AD, meeting BC, BD at F, G ; shew that FG is parallel to CD. 9. From any point in the base of a triangle, straight lines are drawn parallel to the sides: shew that the intersections of the diagonals of every parallelogram so formed lie in a straight line. 10. If the sides of a triangle be produced, the bisectors of the external angles meet the opposite sides in three points which lie in a straight line. 11. If straight lines bisect the interior and exterior angles at A of the triangle ABC in D and E respectively; prove that BD, BC, BE form an harmonica! progression. 12. The diagonals of a parallelepiped bisect one another. 13. The mean point of a tetrahedron is the mean point of the tetrahedron formed by joining the mean points of the triangular faces ; and also those of the edges. 14. If the figure of Ex. 11, Art. 7, be that of a gauche quadri- lateral (a term employed by Chasles to signify that the triangles EX. 15.] VECTOR ADDITION AND SUBTRACTION. 31 AOD, BOD are not in the same plane), the lines QP, DO, RS will meet in a point, provided AP OS . AQ DR 15. If through any point within the triangle ABC, three straight lines MN, PQ, RS be drawn respectively parallel to the sides A B, AC, BC ; then will MN P RS _ AB 1U JfU~~' 1C. A BCD is a parallelogram; E, the point of bisection of AB ; prove that AC, DE being joined will trisect each other. 17. ABCD is a parallelogram ; PQ any line parallel to CD ; PD, QC meet in S, PA, QB in R prove that AD is parallel to ss. CHAPTER III. VECTOR MULTIPLICATION AND DIVISION. 15. WE trust we have made the reader understand by what we stated in our Introductory Chapter, that, whilst we retain for 'multiplication' all its old properties, so far as it relates to ordi- nary algebraical quantities, we are at liberty to attach to it any signification we please when we speak of the multiplication of a vector by or into another vector. Of course the interpretation of our results will depend on the definition, and may in some points differ from the interpretation of the results of multiplication of numerical quantities. It is necessary to start with one limitation. Whereas in Algebra we are accustomed to use at random the phrases ' multiply by' and 'multiply into' as tantamount to the same thing, it is now impossible to do so. We must select one to the exclusion of the other. The phrase selected is 'multiply into'; thus we shall understand that the first written symbol in a sequence is the operator on that which follows : in other words that a/2 shall read 'a into /?', and denote a operating on /?. 16. As in the Cartesian Geometry, so v here we indicate the position of a point in space by its relation to three axes, mutually at right angles, which we designate the axes of x, y, and * respectively. For graphic representation the axes of x and y are drawn in the plane of the paper whilst that of z being perpendicular to that plane is drawn in perspective only. As in ordinary ART. 17.] VECTOR MULTIPLICATION AND DIVISION. 33 geometry we assume that when vectors measured forwards are represented by positive symbols, vectors measured backwards will be represented by the corresponding negative symbols. In. the figure before us, the positive directions are forwards, upwards and outwards; the corresponding negative directions, backwards, downwards and inwards. With respect to vector rotation we assume that, looked at in. perspective in the figure before us, it is negative when in the direction of the motion of the hands of a watch, positive when in the contrary direction. In other words, we assume, as is done in modern works on Dynamics, that rotation is positive when it takes place from y to z, z to x, x to y : negative when it takes place in the contrary directions (see Tait, Art. 65). Unit vectors at riglvt angles to each otJier. 17. DEFINITION. If i, j, k be unit vectors along Ox, Oy, Oz respectively, the result of the multiplication of i into j or ij is defined to be the turning of j through a right angle in the plane perpendicular to i and in the positive direction ; in other words, the operation of i on j turns it round so as to make it coincide with k ; and therefore briefly ij = k. To be consistent it is requisite to admit that if i instead of operating on^' had operated on any other unit vector perpendicular to i in the plane of yz, it would have turned it through a right angle in the same direction, so that ik can be nothing else than j. Extending to other unit vectors the definition which we have illustrated by referring to i, it is evident that j operating on k must bring it round to i, orjk i. Again, always remembering that the positive directions of rotation are y to z, z to x, x to y, we must have ki =j. 18. As we have stated, we retain in connection with this definition the old laws of numerical multiplication, whenever .numerical quantities are mixed up with vector operations; thus 2i . 3j = Gij. Further, there can be no reason whatever, but the contrary, why the laws of addition and subtraction should undergo T. Q. 3 34 QUATERNIONS. [CHAP. III. any modification when the operations are subject to this new definition ; we must clearly have Finally, as we are to regard the operations of this new de- finition as operations of multiplication magnitude and motion of rotation being united in one vector symbol as multiplier, ju*t as magnitude and motion of translation were united in one vector symbol in the last chapter we are bound to retain all the laws of algebraic multiplication so far as they do not give results inconsistent with each other. In no other way can the conclusions be made to compare with those deduced from the corresponding operations in the previous science. Thus we retain what Sir William Hamilton terms the associative law of multiplication : the law which assumes that it is indifferent in what way operations are grouped, provided the order be not changed ; the law which makes it indifferent whether we consider a be to be a x be or ab x c. This law is assumed to be applicable to multiplication in its new aspect (for example that ijk ~ ij . &), and bding assumed it limits the science to certain boundaries, and, along with other assumed laws, furnishes the key to the interpreta- tion of results. The law is by no means a necessary law. Some new forms of the science may possibly modify it hereafter. In the meantime the assumption of the law fixes the limits of the science. The commutative law of multiplication under which order may be deranged, which is assumed as the groundwork of common algebra (we say assumed advisedly) is now no longer tenable. And this being the case it is found that the science of Quaternions breaks down one of the barriers imposed by this law and expands itself into a new field. ij is not equal toji, ib is clearly impossible it should be. A simple inspection of the figure, and a moment's consideration of the definition, will make this plain. The definition imposes on i as an operator on^' the duty of turning^' through a right angle as if by a left-handed turn with a cork-screw handle, thus throwing j up from the plane xy; when, on the other hand,J is the operator ART. 19.] VECTOR MULTIPLICATION AND DIVISION. 35 and i the vector operated on, a similar left-handed turn will bring i down from the plane of xy. In fact ij = Jc, ji = k, and so y = -ft- 19. We go on to obtain one or two results of the application of the associative law. 1. Since ij = k, we have i . ij = ik = j. Now by the law in question, or i = l. Our first result is that the square of the unit vector along Ox is 1 ; and as Ox may have any direction whatever, we have, gene- rally, the square of a unit vector = 1. In other words, the repetition of the operation of turning through a right angle reverses a vector. 2. Again, ijk = i .jk = i . i = i 2 = 1. Similarly it may be proved that jki = kij = -l, or no change is produced in the product so long as direct cyclical order is maintained. 3. But ikj=i . kj = i . i tf = + 1 ; .-. ijk^-ikj, or a derangement of cyclical order changes the sign of the product. This last conclusion is also manifest from Art. 18. Vectors generally not at right angles to each other. 20. We have already (Art. 8) laid down the principle of separation of the vector into the product of tensor and unit vector ; and we apply this to multiplication by the considerations given in Art. 18, from which it follows at once that if a be a vector along Ox containing a units, /? a vector along Oy con- taining b units, a = ai, ft = bj, and a/? = abij. 32 t 36 QUATERNIONS. [CHAP. III. In the same way a 2 = ai . ai = a*i 2 = a 2 , or the square of a vector is the square of the corresponding line with the negative sign. Seeing therefore the facility with which we can introduce tensors whenever wanted, we may direct our principal attention, as far as multiplication is concerned, to unit vectors. 21. We proceed then next to find the product a/3, when a and /3 are vectors not at right angles to one another. 1. Let a, ft be unit vectors. Let OA - a, OB = ft. Take OC = y, a unit vector perpen- dicular to OB and in the plane BOA. Take also DO or DO produced- e, a unit vector perpendicular to the plane BOA. Draw AM, AN perpendicular to OB, OC, and let the angle BO A = ; then vector OA = OM + MA = OM+ ON (Art. 1) = part of OB + part of OC (Art. 3). Now it is evident that OM as a line is that part of OB which is represented by the multiplier cos 6, or OM = OB cos 9, and similarly that ON=OCs\nO: consequently (Art. 3) the same applies to them as vectors ; i. e. vector OM=(3cosO, vector ON=y sin 6} .'. a = (3 cos 6 + y sin 6, and a/3 = (/3 cos + y sin 0) /3 = /3 2 cos + y/3 sin 0. But /3 2 = -l (19. 1), y/3 = e (17); [Observe that y, (3 and c of the present Article correspond toj, i and -k of Art. 17.] .'. aft cos 0+ esintf. ART. 22.] VECTOR MULTIPLICATION AND DIVISION. 37 2. If a, /3 are not unit vectors, but contain To. and Tft units respectively, we have at once, by the principle laid down in Art. 20, a/2 = TaTfi (- cos + e sin 0). 3. It thus appears that the product of two vectors a, /3 not at right angles to each other consists of two distinct parts, a numerical quantity and a vector perpendicular to the plane of a, /?. The former of these Sir William Hamilton terms the SCALAR part, the latter the VECTOR part. We may now write a/3 = Saj3 + Fa/8, where S is read scalar, F vector : and we find 7afi = TaTfi sin 0. 4. The coefficient of e in Fa/3 is the area of the parallelogram whose sides are equal and parallel to the lines of which a, /? are the vectors. 22. To obtain /3a we have, a and (3 being unit vectors, a = /? cos 6 + y sin 6 ; - - cos - e sin (Art. 19. 1 and 18) ; therefore generally (3a = TaTft (- cos - e sin 0). It is scarcely necessary to remark that whilst y operating on ft turns it inwards from OB to DO produced, /? operating on y turns it outwards from 00 to OD, causing it to become - e. We have therefore 1. Sap = S(3a. 2. 7ap = -V{3a. 3. ap + fia = 2Sa/3. 4. ap-p a =2Vap. /1C 38 QUATERNIOXS. [CHAP. III. 5. a + P=a + Pa + P 6. (a-pY = a 3 -2Sap+p 2 . 7. If a, (3 are at right angles to each other, Saj3 = 0, and conversely. 8. Vap is a vector in the direction perpendicular to the plane which passes through a, /?. 9. a*/3* = a/3 . Pa because /3* is a scalar ; af3 - Va0) Note, a 2 ft 2 must not be confounded with (aft) 3 . 23. Before proceeding further it is desirable we should work out a few simple Examples. Ex. 1. To express the cosine of an angle of a triangle in terms of the sides. Let ABC be a triangle ; and retaining the usual notation of Trigonometry, let CB=a, CA=/3; then (vector A) s = (a - /?)' = a s -2Sa.p + p* (22. 6), or, changing all the signs to pass from vectors to lines (20) and applying 21. 3, Ex. 2. To express the relations between the sides and opposite angles of a triangle. Let CB = a, CA = p, JBA = y. Then CB + BA = CA gives . . a* = a (P y) a/3 ay. Take the vectors of each side. AET. 23.] VECTOR MULTIPLICATION AND DIVISION. 30 Xow Fa* = 0, for a 2 = - a 3 has no vector part, i e. (21. 3) abe sin C = ace sin JS, or b sin C c sin B ; Le. b : c :: sinJ5 .: sin (7. Ex. 3. TAe sum of the squares of the diagonals of a paral- lelogram is equal to t/te sum of the squares of the sides. Retaining the notation and figure of Ex. 1, Art. 7, .'. CB* + DA 3 = 2a 2 + 2^, and, changing all the signs, we get (20) for the corresponding lines, Ex. 4. Parallelograms upon the same base and between t/te same parallels are equal. It is necessary to remind the reader of what we have already stated, that examples such as this are given for illustration only. We assume that the area of the parallelogram is the product of two adjacent sides and the sine of the contained angle. Adopting the figure of Euclid I. 35 and writing TVfia. as the tensor multiplier of FySa so as to drop the vector e on both sides; we have, calling LA, a ; BC, ft ; BE=BA+AE .e. a remembering that je/3 2 has no vector part, Hence T.Vfta^T (BC . BE), i. e. BC . BA sin ABC = BC.BE sin EEC (21. 3), which proves the proposition. 40 QUATERNIONS. [CHAP. III. Ex. 5. On the sides AS, AC of a triangle are constructed any two parallelograms ABDE, ACFG : the sides DE, FG are produced to meet in II. Prove that the sum of the areas of the parallelograms ABDE, ACFG is equal to the area of the parallelogram whose adjacent sides are respectively equal and parallel to BC and AH. Let BA = a, AE=$, AC = y, GA=S, then AH = (3 + xk, and AH= $yy; .: VaAH = Vap and VyAH=-VyS = V8y(22. 2), hence F (a + y) AH= Vap + FSy, i. e. (21. 4), the parallelogram whose sides are parallel and equal to BC, AH, equals the two parallelograms whose sides are parallel and equal to BA^ AE $ GA, A C respectively. [The reader is requested to notice that the order GA, AC is the same as the order BA, AE, and BA, All : so that the vector e is common to all.] Ex. 6. If be any point whatever either in the plane of the triangle ABC or out of that plane, the squares of the sides of the triangle fall short of three times the squares of the distances of the angular points from 0, by the square of three times the distance of the mean point from 0. Let OA = a, OB = p, OC = y, then (Art. 14), OG = \ (a + p + y), o or a* + (l 2 + y a + 2S(ap + l3 7 + ya) = 30G*. Now AB=p-a, C = y-p, CA=a-y, . '. AB 2 + BC 1 + CA 2 = 2 (a 2 + yS 2 + y 2 ) - 2S (a + /3y + ya) and the lines AB* + BC 2 + CA 3 = 3 (OA* + OB 2 + OC 2 ) - (30G)*. Ex. 7. The sum of the squares of the distances of any point from the angular points of the triangle exceeds the sum of the ART. 23.] VECTOR MULTIPLICATION AND DIVISION. 41 squares of its distances from the middle points of the sides by the sum of the squares of half the sides. Retaining the notation of the last example, and the figure of Ex. 4, Art. 7, OZ) = l(/3 + y) > OE= l -(y + a\ 0^=1 ( + ); .'. 4 (OD* + OE 2 + OF 9 ) = 2 (a 2 + /3 2 + y 2 ) + 2S (a/? + Py + ya) = 4 (a 2 + ft 2 + y 2 ) - (AB 2 + BC 2 + CA 2 ) ; A 7? 2 4- T?^ 8 -i- ^y4* .-. as lines OD 2 + OE 2 + OF 1 + + oyi = ^^ + O^ 2 + (9C 12 . Ex. 8. IVie squares of the sides of any quadrilateral exceed the squares of the diagonals by four times the square of the line which joins the middle points of the diagonals. Retaining the figure and notation of Ex. 8, Art. 7, we have squares of sides as vectors and squares of diagonals therefore the former sum exceeds the latter by Tlierefore as lines the same is true. Note. The points A, B, C, D need not be in one plane. QUATERNIONS. [CHAP. in. Ex. 9. Four times the squares of the distances of any point whatever from the angular points of a quadrilateral are equal to thi sum of the squares of the sides, the squares of the diagonals and the square of four times the distance of the point from the mean point of t/ie figure. With the notation of Art. 14, and the figure of Ex. 7, Art. 7> we have squares of the sides + squares of the diagonals - 3 (a 2 + ft 2 + y 2 + o 2 ) - 2S (a/3 + ay + aS + /3y + 5 + yS). Now (Art. 14) (a + /? + y + S) 2 = (WX) 2 j . '. (4CLr) 8 + squares of sides -f squares of diagonals = 4 (OA* + OB 2 + OC 2 + OD*). Ex. 10. The lines which join the mean points of three equila- teral triangles described outwards on the three sides of any triangle form an equilateral triangle whose mean point is the same as that of the given triangle. Let P, Q, R be the mean points of the equilateral triangles on BG, CA, AB; PD= a, DC - (3, CE = y , EQ = 8 ; and let the sides of the triangle ABC be 2a, 26, 2c. ft 2 -f ART. 23.] VECTOR MULTIPLICATION AND DIVISION. 43 Changing all the signs and observing that 2 /Sa(3 0, Say = -- -p; ab sin C, &c. V we have (writing the results in the same order), Q 2 = ~ + a 2 + b s + ^ + o o 22 2 + . ab sin C + -= ab cos C 2ab cos C + - . ab sin C + V'" " v " 4 4 = K ( 2 + 6 2 - ab cos C) + 7^06 sin C > vi = | (a 2 + 6 2 + c 2 ) + ~ area of ^ BC, o Jo which being symmetrical in a, b, c proves that PQR is equilateral. Again, G being the mean point of ABC, i T 7J/-V2 a 2 a 2 46 2 4 _, 4 . and line PG =-^- + -& + - - + ^ 7^ a6 sin C - r a5 cos C o y y o ,y/o y - (a 2 + 6 2 + c 2 ) + area and 6f is the mean point of the equilateral triangle PQR. Ex. 11. In any quadrilateral prism, tlie sum of the squares of the edges exceeds the sum of the F , squares of the diagonals by eight times the square of the straight line which joins the points of inter- section of t/ie two pairs of diagmials. sum of squares of edges = 2 {a 2 + 2 + (y- a) 2 + (y - /5) 2 + 28 2 } = 2 (2a 2 + 2 ft 3 + 2y 2 + 28 2 - 2Say 44 QUATERNIONS. [CHAP. III. sum of squares of diagonals = (S + 7 ) 2 + (S- 7 ) 2 +(S + a-y8) 2 + (S + /3-a) 2 = 2 {a 2 + ft 3 + / + 2S 2 - 2Sa/3}. Also 10Gf = l(8 + y ) = vector to the point of bisection of CD, and therefore to the point of intersection of OG, CD, and vector from to the point of bisection of AF, as also to that of BE, and therefore to the intersection of A F, BE hence vector which joins the points of intersection of diagonals eight times the square of this vector = 2 (a 2 + P 2 + / + 2Sap - 2Say - 2Spy), which, added to the sum of the squares of the diagonals, makes up the sum of the squares of the edges. 24. DEFINITION. We define the quotient or fraction , where a and p are unit vectors, to be such that when it operates on a it produces p or . a = /?. This form of the definition enables us to strike out a by a dash made in the direction of ordinary writing, thus . a = p. is therefore that multiplier which, operating a a. on a, or on p cos + y sin (21), produces p. Now cos + e sin operating on p cos + y sin produces P cos 8 + (y + e/3) sin cos + ey sin 2 0. But a glance at the figure (Art. 21) will shew that and ART. 25.] VECTOR MULTIPLICATION AND DIVISION. 45 .-. cos 6 + e sin operating on /? cos + y sin produces /3 ; hence = cos + c sin 0. a It may be worth while to exhibit another demonstration of this proposition : thus - . a/2 = (3 . ft (by the associative law) = - 1 . (19 . 1). i.e. (21 . 1) . (-cos0 + esin0) = -l. Now (cos 6 + e sin 6) ( cos + e sin 6) = - cos 2 sin 2 i . ~ A > .*.= cos + e sin 0. a COR. = -a(by 22). 25. 1. DEFINITION. Still retaining a, (3 as unit vectors, since operating on a causes it to become /3, it may be defined as a VERSOR acting as if its axis were along OD (Fig. Art. 21). By comparing the result of that article with the definitions of Art. 17, it is clear that or cos + c sin is an operator of the same character as k or e (as we have now called the corresponding unit vector) ; with this difference only, that whereas k or c as an operator would turn a through a right angle, cos + e sin 9 turns it, in the same direction, only through the angle : cos 6 + e sin 6 is then the versor through the angle 0. 2. If a, ft are not unit vectors, the considerations already advanced render it evident that TB Now j~- is itself of the nature of a tensor, for it is a numerical J.O. quantity, hence - is the product of a tensor and a versor. 46 QUATERNIONS. [CHAP. III. 26. By comparing the last Article with Art. 22 it appears that generally the product or quotient of two vectors may be expressed as the product of a tensor and a versor. This product Sir W. Hamilton names a QUATERNION. COR. It is evident that a quaternion is also the sum of a scalar and a vector. 27. (1) If a > A 7 are un it vectors in the same plane, c a unit vector perpendicular to that plane ; we have seen that - operating on a turns it round about e as an axis to bring it into the position /?. If now - be a second operator about the same axis in the same direction acting on (3, it will bring it into the position y. But it is evident that - acting on a would at once have brought it into the position a y. This is equivalent to the fact that ^ . = - ; or in another pa a form (Art. 24) that (cos < + c sin </>) (cos 9 + c sin 9} = cos (9 + </>) + c sin (9 + </>). Prom this it is evident thnt the results of Demoivre's Theorem apply to the form cos 9 + c sin 9. Further, it is evident that since cos 9 + e sin 9 operating with c as its axis, turns a vector through the angle 9, whilst e itself acting in the same direction turns it through a right angle, cos 9 + c sin 9 is part of the operation designated by e, viz. that part which bears to the whole the proportion that 9 bears to a right angle. (2) Remembering then that the operations are of the nature of multiplication, it becomes evident that cos 9 + c sin 9 as an ~ 29 operator may be abbreviated by or e w . And since (cos 9 + e sin 9} (cos < + e sin <) = cos (9 + <) + e sin (0 + <), ART. 28.] VECTOR MULTIPLICATION AND DIVISION. 47 we shall have or the law of indices is applicable to this operator. (3) Now we have already seen (19. 1) that c 2 = 1 ; .'. 4 = + l. Conversely, if c" = e, n must be an odd number; if e" = -l, n must be an odd multiple of 2 ; and if c" = + 1, n must be an even multiple of 2. (4) "When a, (3 are not units, the introduction of the corre- sponding tensor can be at once effected. We conclude that a quaternion may be expressed as the power of a vector, to which the algebraic definition of an index is applicable. 28. Reciprocals of quaternions unit vectors. 1. Since a.a = o s = 1, and -.a=l (Def. Art. 24) = a . a ', .: - = a, or a" 1 =-a: a or the reciprocal of a unit vector is a unit vector in the opposite direction. 2. Again, a.- = a(-a) = l=-.a; a a or a vector is commutative with its reciprocal. 3. If q be a versor ( say cos + e sin 0, or - J , - . q = 1 (Def. extended). Now = q ; a. .'.ft- qa, by operating on a. 48 QUATERNIONS. [CHAP. III. a 1 Also -^ = - , a = - /?, by operating on {3, and /3 = g-a = q . - (3 ; 1 1 . '. q . - = 1 = - . q, q q ' or q and - are commutative. This is perhaps better demonstrated by observing that ~ ' a~ a~~ > a p p or that if = cos + e sin 6, a then must -= = cos Q e sin 6 ; factors which are from their very nature commutative. Asa verification, we have .75= (cos 6 + sin 6) (cos - e sin 6) a }3 = (cos e) 2 - c 2 (sin ey because e 2 = - 1 (28. 1). When the versors are not units the tensors can be introduced as mere multipliers without affecting the versor conclusions. 29. We present one or two examples of quaternion division. Ex. 1. To express sin (0 + <) and cos (0 + <) in terms of sines and cosines of 6 and <. a, /?, y being unit vectors in the same plane (Fig. Art. 27), we have - = cos + c sin 6, a. ART. 29.] VECTOR MULTIPLICATION AND DIVISION. 49 y . jr = cos <f> + e sin <f>, 2 = cos (0+ <f>) + sin (&+ <). But ?=!.; a (3 a . . cos (5 + <) + 6 sin (# + <) = (cos + e sin 0) (cos < + e sin <) ; whence multiplying out and equating, we have sin (6 + <) = sin 6 cos < + cos 6 sin <, cos (6 + <) = cos cos < sin 6 sin <. COR. If the action of the versors be in opposite directions, y3 lying beyond y, \ve have (Art. 28) - = cos (0 -<}>)- sin (0 - <). But - = cos d> + e sin d>. y -~ = cos - e sin ; p* a a B . ' - -7i - Rives y 3 y * cos (0 - <) - sin (0 - <) = (cos - sin 6) (cos + e sin <), whence sin (0 <) = sin cos ^> cos sin <f> , cos ($-<{)) = cos cos /> + sin 5 sin <. Ex. 2. To ^%^ tlte cosine of the angle of a spherical triangle in terms of the sides. Let a, (3, y be unit vectors OA, OS, OC not in the same plane, then . i.e. taking the scalar of each side, a fvP v a \ - cos a ~ cos c cos 6 + o ' . ( V - . V - J B T. Q. . T. Co 50 QUATERNIONS. [CHAP. III. Now /SV V is sin c sin b x cosine of the angle between a y perpendiculars to the planes AD, AC, and is therefore sin b sin c cos A ; /. cos a = cose cos b + sin c sin b cos A. The reader will observe that in accordance with the results of Art. 21, the sign of the term involving cos .4 is +, seeing that it is in fact cosine (supplement of A). Ex. 3. The angles of a triangle are together equal to two right angles. What we shall prove in fact is that the exterior angles formed by producing the sides in the same direction are equal to four right angles. Let unit vectors along BC, CA, AB be a, /?, y ; and let the exterior angles formed by producing BC, CA, AB be 0, </>, i/^; then 29 e"a = /3(27. 1), 24> 29 2 .'. e"" . t.* a = c' r 2<ft 2j 29 2j and e* 1 . e* . "' a = ' r 2^ 2^1 29 so that " . e 17 . e 71 " = 1, = 1 (27. 2). 2 Hence (27. 3), - (0 + < + iff) is an even multiple of 2. The first value is 4 ; or the exterior angles of a triangle are equal to four right angles. ART. 29.] VECTOR MULTIPLICATION AND DIVISION. 51 It will be seen that the demonstration here given is of the nature of that given by Prof. Thomson in the Notes to his Euclid. i [More directly From these or A + + = ir.] Ex. 4. In the figure of Euclid i. 47 the three lines AL, BK, GF meet in a point. Let BC = a, CA = /?, AB = y; the sides being as usual denoted by a, b, c. Let i be the vector which turns another negatively through a right angle in the plane of the paper, so that If BK, AL meet in 0, BO and BO x (a + iff) = y + yia, xSa (a. + ifi) = - Say, Say etc cos B ' - - - - Sa (a + if}) a 2 + ab sin c> and xSa/3 = ySia/3 b be 42 52 QUATERNIONS. [CHAP. III. which being symmetrical in b and c shews that CF, AL intersect in the same point in which BK, AL intersect. BO _ c 3 C*OR. feince ,, fi BK CO we have also CF a*+bc' AO bo BD ~ a" + be ' AO^ B0_ CO _ c a + b* + be '' BD + ~BK + CF~ a* + bc Ex. 5. If A BCD be a quadrilateral inscribed in a circle ; Let unit vectors along AB, BC, CD, DA be a', ft', y, X ; and let the exterior angles at B and D be 6 and < respectively ; then a'py = (- cos 6 + e sin 6) y' (21. 1) = (cos (ft + e sin = 8' (25. 1); therefore, introducing the tensors, Conjugate Quaternions. 30. If we designate by y the expression cos + e sin 0, we have seen that it may be regarded as a versor through an angle in a certain direction. Now if we write in place of 6 in this expression it assumes the form cos c sin 0, which must on the same hypotheses be regarded a versor through the angle 6 in the contrary direction. When the quaternion is completed by the introduction of a tensor Tq, if we retain the same tensor to both forms of the ART. 30.] VECTOR MULTIPLICATION AND DIVISION. 53 versor, we have Sir "W. Hamilton's conjugate quaternion defined thus : The conjugate of a quaternion q, written Kq, has the same tensor, plane and angle as q has, only the angle is taken in the reverse way. The analogy between q and Kq is precisely the same as that which exists between the two forms R (cos </> + - 1 sin <) and R (cos < J - 1 sin <) ; and as the product of the latter form is R 2 , so the multiplication of the former produces (Tqf. If we put q = Sq + Vq, we shall have Kq = Sq Vq, and qKq = (Sq)' + (TVq)' t for (Vqf = -(TVqY, Art. 20. It is almost self-evident that, since the change of order of multiplication of two vectors produces no other change than that of the sign of the vector part of the product (22), q and r occurring in a changed order. The following is a demonstration, Let q=Tq( cos + a sin 0), r = Tr( cos < + ft sin <), a and ft being unit vectors ; then qr = TqTr (cos cos <f> a sin 6 cos < - (3 cos 6 sin < + aft sin sin <), KrKq = TqTr (- cos < - (3 sin <) (- cos - a sin 0) = TqTr (cos 6 cos < + a sin 6 cos < + ft cos sin <f> + fta sin sin <). Now observing that /3a has the same scalar part with aft, but the vector part with a contrary sign, we see that the two ex- 5'4 QUATERNIONS. [CHAP. III. pressions for qr and for KrKq likewise have the same scalar part, but that their vector parts have contrary signs. Hence K (qr) = KrKq. (See Tait, 79 et sq.) 31. We propose, in this Article, to give and interpret one or two formulae, relating to three or more vectors, which are indis- pensable to our progress, reserving to a separate Chapter the demonstration and application of other formulae, the value of which the reader can hardly as yet be expected to understand. 1. To express S . a/3y geometrically. First suppose a, ft, y to be unit vectors OA, OB, 00. Let AOB-Q, and the angle which 00 makes with the plane A OB = (ft ; then since aft = - cos 6 + e sin 6 (Art. 21), where e is perpendicular to the plane A OB, S . afty = S(- cos + e sin 6) y = /Scy sin 6. Now Sey = cos . angle between e and y = sin . angle between plane AOB and 00 = sin <f> .'. S. afty= sin <f) sin 0. 0< Next if a, ft, y are not units, but have re- spectively the lengths Ta, Tft, Ty, or a, b, c; we shall have S . afty abc sin sin $. But db sin is the area of the parallelogram of which the adjacent sides are a, b ', and c sin < is the perpendicular from C on the plane of the parallelogram ; . '. S . afty = db sin 6 . c sin < = volume of parallelepiped of which three con- terminous edges are OA, OB, 00. ART. 31.] VECTOR MULTIPLICATION AND DIVISION. 55 2. From the nature of the case, no change of order amongst the vectors a, /?, y can make any change in the value (apart from the sign) of the scalar of the product of the three vectors ; for it will in every case produce the volume of the same parallelepiped. .-. S.aj3y = S .yap = S.ay/3, &c. COR. 1. The volume of the triangular pyramid, of which OA, OB, OC are conterminous edges is -^S . a/2y. COR. 2. If a, /?, y are in the same plane, </> = ; .'. S.ay = 0. Conversely, if S . a/?y = 0, none of the vectors a, /?, y being themselves 0, we must have either or <f> = ; hence in either case the three vectors are co-planar. 3. Since Fa/3 = y' (21. 3), a vector perpendicular to the plane OAB (fig. of formula 2) ; F/?y = a', a vector perpendicular to the plane OBC ; and since y, a! are both perpendicular to 0, the line along which is the vector (3 ; OB is perpendicular to the plane which passes through y', a', and therefore (21. 3) is in the direction of Vy'a ; hence V ( Vafi Vfty) = Vy'a - mfi, or the vector of the product of two resultant vectors, one of the constituents of each of which is the same vectoi 1 , is a multiple of that vector. 4. If OA = a, 0.8 = (3, OJ) = S, OE=e; and if the planes OAB, ODE intersect in OP; it follows, as in (3), that, Vafi and FSe being both perpendicular to OP, V(VapVSe) is along OP and is therefore =nOP. 5. Connection between the representation of the position of a point by a vector and its representation by Cartesian co-ordinates. If a?, y, z be the perpendicular distances of a point P in space from the planes of yz, zx, xy respectively (fig. of Art. 16); *, j, k 56 QUATERNIONS. [CHAP. III. unit vectors in the directions of x, y, z ; then xi is the vector of which the line is x (Art. 3) ; consequently OM along Ox, MN parallel to Oy and NP parallel to Os, being x, y, z as co-ordinates, they are xi, yj, zk as vectors. Now vector OP = OM+ MN+ NP, and is therefore p = xi + yj + zk. The same method of representation is evidently applicable when the planes of reference are not mutually at right angles. If x, y, z be the co-ordinates of P referred to oblique co-ordinates; a, /?, y unit vectors parallel respectively to x, y, z ; then vector OP = xa + y{3 + zy. COR. When x, y, z are at right angles to one another, p = xi + yj + zk gives Sip = -x, Sjp = -y, Skp = -z; .-. (Sip)' + (Sjp)> + (Sk P )* = x' + y* + z* = OP*. Ex. To find the volume of the pyramid of which the vertex is a given point and the base the triangle formed by joining three given points in the rectangular co-ordinate axes. Let A, B, C be the three given points ; x, y, z the co-ordinates of the given point P, then vector OA = ai, OB = bj, OC-ckj and OP = xi + = -{xi + yj+(z-c)k}. ART. 31.] VECTOR MULTIPLICATION AND DIVISION. 57 Now the volume of the pyramid PASO is ~S(PA.PB.PC} (31. 2. Cor. 1) = - ^ S . { (x - a) i + yj + zk} {xi + (y - b) j + zk} [xi + yk + (z - c) k}. Multiplying out and observing that only terms which involve all of the three vectors i, j, k produce a scalar in the product, we get (+ or -) Yol. = - ^ {(x - a) (bz + cy be) - cxy bxz} x y z n \ -+^ + 1). a o c J 1 fx y z ^A>. i i ^ i "6 The sign of the result will of course depend on the position P. ADDITIONAL EXAMPLES TO CHAP. III. 1. If in the figure of Euclid i. 47 DF, GH, KE be joined, the sum of the squares of the joining lines is three times the sum of the squares of the sides of the triangle. The same is true whatever be the angle A. 2. Prove that AD* (Art. 7, Ex. 4) = 2 (AB* + AC*)- BO 3 . 3. If P, Q, R, S be points in the sides AB, BC, 02), DA of a rectangle, such that PQ US, prove that AH* + OS' = AQ* + OP 2 . 4. The sum of the squares of the three sides of a triangle is equal to three times the sum of the squares of the lines drawn from the angles to the mean point of the triangle. 58 QUATERNIONS. [CHAP. III. 5. In any quadrilateral, the product of the two diagonals and the cosine of their contained angle is equal to the sum or difference of the two corresponding products for the pairs of opposite sides. 6. If a, b, c be three conterminous edges of a rectangular parallelepiped ; prove that four times the square of the area of the triangle which joins their extremities is 7. If two pairs of opposite edges of a tetrahedron be respect- ively at right angles, the third pair will be also at right angles. 8. Given that each edge of a tetrahedron is equal to the edge opposite to it. Prove that the lines which join the points of bisection of opposite edges are at right angles to those edges. 9. If from the vertex of a tetrahedron OABG the straight line OD be drawn to the base making equal angles with the faces OAB, OAC, OBC ; prove that the triangles OAB, OAC, OBG are to one another as the triangles DAB, DAG, DBG. CHAPTER IV. THE STRAIGHT LINE AND PLANE. 32. EQUATIONS of a straight line. 1. Let ft be a vector (unit or otherwise) parallel to or along the straight line: a the vector to a given D A P point A in the line, p that to any point what- ever P in the line, starting from the same origin ; then AP is a vector parallel to /3 = x{3, say, and OP = OA + AP gives p = a + x(3(I) as the equation of the line. 2. Another form in which the equation of a straight line may be expressed is this : let A a, OS = (3 be the vectors to two given points in the line ; then Of course the ft of No. 2 is not that of No. 1. The first form of the equation supposes the direction of the line and the position of one point in it to be given, the second form supposes two points in it to be given. 3. A third form may be exhibited in which the perpendicular on the line from the origin is given. 60 QUATERNIONS. [CHAP. IV. Let OD perpendicular to AP = 8; then because OD is perpendicular to AP (22. 7) ; Le.S&p = C (3), where C is a constant. (Note. In addition to this we must have the equation of the plane of the paper, in which p is tacitly supposed to lie. This may be written as Sep = 0.) 33. Equation of a plane. Let P be any point in the plane, OD perpendicular to the plane ; and let OD = S, OP = p- } then p - 8 = DP, which is in a direction perpendicular to OD ; or COR. 1. If SBp = C be the equation of a plane, 8 is a vector in the direction perpendicular to the plane. Con. 2. If the plane pass through 0, p can have the value zero, . '. SBp = is the equation. COR. 3. Since a vector can be drawn in the plane through Z>, parallel to any given vector in or parallel to the plane ; if ft be any vector in or parallel to the plane, SS/3 = 0. 34. We proceed to exhibit certain modifications of the equations of a straight line and plane, and one or two results immediately deducible from the forms of those equations. 1. To find the equation of a straight line which is perpen- dicular to each of two given straight lines. Let /?, y be vectors from a given point A in the required line, and parallel respectively to the given lines. ART. 34.] THE STRAIGHT LINE AND PLANE. 61 If OA = a as before, then since (22. 8) F/3y LS a vector along the line whose equation is required ; we have p a = x T /7 /3y, or p = a + x F/3y, as the equation of the line. 2. To find the length of the perpendicular from the origin on a given line. Equation (1) of Art. 32 is p - a. + x(3. If now p = OD = 8 ; we get S&* = SSa, or -OD* = SSa US being the unit vector perpendicular to the line. COR. The same result is true of a plane. 3. To find the length of the perpendicular from a given point on a given plane. Let Sap = C be the equation of the plane, y the vector to the given point. Then if the vector perpendicular be xa (33. Cor. 1), p = y 4- xa gives Say + xa* = (7, and the vector perpendicular is xa = + a" 1 (C - Say) ; the square of which with a sign is the square of the perpendi- cular. 4. To find the length of the common perpendicular to each of two given straight lines. 62 QUATERNIONS. [CHAP. IV. Let (3, /3 l be unit vectors along the lines ; a, a x vectors to given points in the lines ; p = a + x(3, Pi = !+, A , the vectors to the extremities of the common perpendicular 8. Then since 8 is perpendicular to both lines, it is perpendicular to the plane which passes through two straight lines drawn pa- rallel to them through a given point ; But 8 = p - pj = a + x(3 aj - hence S . B/3/3, = S . (a - a,) i. e. S (y 7(3(3, . ft (3,) = S . (a - a,) because . whence 8 = /V is known. 5. To find the equation of a plane which passes through three given points. Let a, ft, y be the vectors of the points. Then p a, a (3, (3 y are in the same plane. .-. (Art. 31. 2. Cor. 2) S. 0>-a)(a-/8)(-y) = 0, or Sp(Va(3+V(3y+Vya)-S.a(3y = is the equation required. COR. Fa/3 + V(3y + Vya is a vector in the direction perpen- dicular to the plane; therefore (No. 3) the perpendicular vector from the origin = S.a(3y.(Va(3+ V(3y + Fya)' 1 . 6. To find the equation of a plane which shall pass through a given point and be parallel to each of two given straight lines. ART. 34.] THE STRAIGHT LINE AND PLANE. 63 Let y be the vector to the given point, p = a + xft, p = a l + a; 1 /8 1 the lines ; then if lines be drawn in the required plane parallel to each of the given straight lines these lines as vectors will be ft, ft 1 : also p y is a vector line in the plane ; .'. S.ftft 1 (p-y) = Q (31. 2. Cor. 2), which is the equation required. 7. To find the equation of a plane which shall pass through two given points and be perpendicular to a given plane. Let a, ft be the vectors to the given points, SSp C the equa- tion of the plane ; then the three lines p a, a ft, 8 are vectors in the plane ; or .pa 8. To find the condition that four points shall be in t/te same plane. 1. Let OA, OS, 00, OD or a, ft, y, 8 be the vectors to the four points ; then 8 a, 8 ft, 8 y are vectors in the same plane ; .-. S . (8 - a) (8 - ft) (8 - y) = (31. 2. Cor. 2), or S.ofty + S.a$y + S.aftS = S.afty (1). 2. Another form of the condition is to be obtained by as- suming that dS + cy + bft + aa = (2), and substituting in equation (1) the value of 8 deduced from this equation. The result is a o c - - = Q (3). Equation (1), or the concurrence of equations (2) and (3) is the condition necessary and sufficient for coplanarity. 9. To find the line of intersection of two planes through the origin. 64 QUATERNIONS. [CHAP. IV. Let Sap = 0, Sftp = be the planes. Since every line in the one plane is perpendicular to a ; and every line in the other perpendicular to ft; the line required is perpendicular to both a and ft, and is therefore parallel to Fa/?, or p = xVaft is the equation. 10. The equation of the plane which passes through and the line of intersection of the planes /Sap = a, Sftp = b is Sp(aft-ba) = 0. For 1 it is a plane through ; 2 if p be such that Sap = a, then must Sftp = b. 11. To find the equation of the line of intersection of the two planes. Let p = ma + nft + xVaft be the equation required. Then Sap = ma? + nSaft = a, since Vaft is perpendicular to a, and similarly aft*-bSaft bSaft-aft* ~ a*ft*-(Saft)* ~ (Vaft)* aSaft-ba 2 aSaB-ba* (Saft)*-a 2 ft 2 ~ (Vaft) 1 35. We offer a few simple examples. Ex. 1. To find the locus of the middle points of all straight lines which are terminated by two given straight lines. Let AP, BQ be the two given straight lines, unit vectors parallel to which are ft, y; AB the line which is perpendicular to both AP, BQ. Let be the middle point of AB; vector A = a ; R the middle point of any line PQ, rector OR = p ', then ART. So.] THE STRAIGHT LINE AND PLANE. 65 But hence, since Sa/3 = 0, Say = 0, Sap = is the equation required ; and the locus is a plane passing through (33. Cor. 2), and perpendicular to OA (33. Cor. 1). Note that, if /? || y, we have simply 2p = x'(3; and, as there is now but one scalar indeterminate, the locus is a straight line instead of a plane. Ex. 2. Planes cut off, from, the three rectangular co-ordinate axes, pyramids of equal volume, to find the locus of the feet of per- pendiculars on them from the origin. Here the axes are given, so that i,j, k are known unit vectors. Let ai, bj, ck be the portions cut off from the axes by a plane, the perpendicular on which from the origin is p. Then p ai is perpendicular to p ; or p = Similarly, p 2 p 2 = cSkp. Hen ce p 6 = abc Sip Sjp,Skp = CSipSjpSkp, since abc is by the problem constant. If x, y, z be the co-ordinates of p this equation gives at once as the equation required. T. Q. 66 QUATERNIONS. [CEAP. IV. Ex. 3. To find the locus of the middle points of straight lines terminated by two given straight lines and all parallel to a given plane. Retaining the figure and notation of Ex. 1, let 8 be the vector perpendicular to the given plane : we have Now SBQP = (33. Cor. 3); 2SoS S/3S and 2p = xp + r-^ y + x ~~ y oyo oyo where a ~ , 6 = ^;- are constants; (oyd for instance is the oyo oyo negative of the cosine of the angle between one of the given lines and the perpendicular to the given plane). Now (B + by is a known vector lying between /3 and y ; call it e, and 2p = ay + xe is the equation required; which is that of a straight line, not generally passing through (32. 1). Ex. 4. OA, OB are two fixed lines, which are cut by lint's AS, A'B' so tJutt the area AOJB is constant/ and also the product OA, OA' constant. It is required to find the locus of the intersec- tions of AS, A'B'. Let the unit vectors along OA, OB be a, ft respectively. OA = ma, OA' m'a, then the conditions of the problem are mn = m'n' = C, mm' = a. ART. 35.] THE STRAIGHT LINE AND PLANE. 67 Now if A, A'B' intersect in P, and OP = p, we have P =OA + AP = ma + x (nj3 ma), p = OA' + A'P m'a + x' (rift m'a) ; or p = ma + xl B /G \ p = m'a + x' ( -. B m'a } ; \m r J x x' and = > . m m AM Hence x = m + m m + a and p = 5 (aa + CS), m* + a^ and the locus required is a straight line, the diagonal of the parallelogram whose sides are aa, Cf$. Ex. 5. To find the locus of a point such that tlie ratio of its distances from a given point and a given straight line is constant all in one plane. Let S be the given point, DQ the given straight line, SP = ePQ the given relation. Let vector SD = a,SP = p, DQ = yy, y being the unit vector along DQ, PQ = xa; then T P = eT(PQ), 52 68 QUATERNIONS. [CHAP. IV. gives p 2 = e 2 PQ 2 , where PQ is a vector, = e* (xa)" But . \ Sap + xa a = a 2 , for Say = ; and a?of=(cL*-Sap) a ; hence a'p' = e* (a 2 - Sap) 2 , a surface of the second order, whose intersection with the plane S . ayp = is the required locus. Ex. 6. TJie same problem when the points and line are not in the same plane. Retaining the same figure and notation, we see that PQ is no longer a multiple of a ; but and because PQ is perpendicular to DQ and p 9 = e 2 (a ySyp p) 2 , a surface of the second order. COR. If e = 1, and the surface be cut by a plane perpendicular to DQ whose equation is Syp = c, the equation of the section is' another plane, so that the section is a straight line. Ex. 7. To find the locus oftJie middle points of lines of given length terminated by each of two given straight lines. ART. 35.] THE STRAIGHT LINE AND PLANE. 69' Retaining the figure and notation of Ex. 1, and calling RP c, we have 2p = xp + yy (1), and 2RP = RP-JtQ=2a + xj3-yy (2). From equation (1) we have Sap = (22. 7), because (3 is a unit vector, 2Syp = xS(3y y. The first of these three equations shews that p lies in a plane through perpendicular to AB (33. Cor. 2). The second and third equations give 2(Sf3p+S{lySyp) Now (2) gives, by squaring, - 4c 2 = 4a 2 + x*( in which, if the values of x and y just obtained be substituted, there results an equation of the second order in p. Hence the locus required is a plane curve of the second order, or a conic section, which by the very nature of the problem must be finite in extent and therefore an ellipse. Ex. 8. If a plane be drawn through the points of bisection of two opposite edges of a tetrahedron it will bisect the tetrahedron. Let D, E be the middle points of OB, AC: DFEG the cutting plane: OA, OB, OC = a, ft, y respectively. OG = my, AF=n((3-a}. The portion ODGEA consists of three tetrahedra whose common vertex is 0, and bases the triangles AEF, EFG, FGD. Kow OE= l - + a 70 QUATERNIONS. [CHAP. IV 00 -I* OG=my, OF=a + n(p-a); and 6 times the volume cut off + S. TJ (a + y) my {a + n ((3 - a)} P (31.2 Cor. --{n + nm + (1 - n) m} S . ay ft . ay/3. But since E, G, D, F are in one plane, and 2m (1 - ) OE - (1 - n) OG + 2mnOD - mOF= 0, we must have (34. 8) 2m (1 - n) - (1 - n) + 2mn - m = ; .'. m + n = 1 ; and 6 times the whole volume cut off = jr of 6 times the whole volume, t hence the plane bisects the tetrahedron. COR. The plane cuts other two edges at F and G, so that AF_ OG_ AE + OC EX. 1.] THE STRAIGHT LINE AND PLANE. 71 ADDITIONAL EXAMPLES TO CHAP. IV. 1. Straight lines are drawn terminated by two given straight lines, to find the locus of a point in them whose distances from the extremities have a given ratio. 2. Two lines and a point S are given, not in one plane ; find the locus of a point P such that a perpendicular from it on one of the given lines intersects the other, and the portion of the perpendicular between the point of section and P bears to SP a constant ratio. Prove that the locus of P is a surface of the second order. 3. Prove that the section of this surface by a plane perpen- dicular to the line to which the generating lines are drawn pei'pen- dicular is a circle. 4. Prove that the locus of a point whose distances from two given straight lines have a constant ratio is a surface of the second order. 5. A straight line moves parallel to a fixed plane and is ter- minated by two given straight lines not in one plane ; find the locus of the point which divides the line into parts which have a constant ratio. 6. Required the locus of a point P such that the sum of the projections of OP on OA and OB is constant. 7. If the sum of the perpendiculars on two given planes from the point A is the same as the sum of the perpendiculars from B, this sum is the same for every point in the line AB. 8. If the sum of the perpendiculars on two given planes from each of three points A, B, C (not in the same straight line) be the same, this sum will remain the same for every point in the plane ABC. 9. A solid angle is contained by four plane angles. Through a given point in one of the edges to draw a plane so that the sec- tion shall be a parallelogram. 72 QUATERNIONS. [CHAP. IV. 10. Through each of the edges of a tetrahedron a plane is drawn perpendicular to the opposite face. Prove that these planes pass through the same straight line. 11. ABC is a triangle formed by joining points in the rect- angular co-ordinates OA, OB, OC ; OD is perpendicular to ABC. Prove that the triangle AOB is a mean proportional between the triangles ABC, ABD. 12. VapVfip + ( Fa/3) 2 = is the equation of a hyperbola in p, the asymptotes being parallel to a, (3. CHAPTER V. THE CIRCLE AND SPHERE. 36. Equations of the circle. Let AD be the diameter of the circle, centre (7, radius = a, P any point. If vector CD = a, CP = p, we have p 2 = a 2 (1). A If however AP = p, we have (p-a) 2 = -a 2 If be any point, . (2). we have (p~y) 2 ? (3)- These are the three forms of the vector equation. Form (2) may be written If OC = c, form (3) may be written -a. EXAMPLES. 37. Ex. 1. Tlve angle in a semicircle is a right angle. Taking the second form p 3 - 2Sap = 0, we may again write it 74- QUATERNIONS. [CHAP. V. therefore p, p 2a are vectors at right angles to one another. But p - 2a is DP ; .. DP A is a right angle. Ex. 2. If through any point within or ivithout a circle, a straight line be drawn cutting the circle in the points P, Q, the pro- duct OP . OQ is always the same for that point. The third form of the equation may be written, (TpY + 2TpS 7 Up + c* - a? = 0, which shews that Tp has two values corresponding to each value of Up, the product of which is c 2 a 2 . Therefore, &c. Ex. 3. If two circles cut one another, the straight line which joins the points of section is perpendicular to tJte straight line which joins the centres. Let 0, C be the centres, P, Q the points of section ; vector OC = a.' } a, b the radii; then (as vectors) .: iSa.OP = C, a constant. Similarly, SaOQ = C, the same constant ; .-. Sa(OQ-OP) = Q > or SaPQ = Q, i.e. PQ is at right angles to 00. Ex. 4. in a fixed point, A B a given straight line. A point Q is taken in the line OP drawn to a point P in AB, such that OP.OQ = k*-, to find the locus of Q. Let OA perpendicular to AB be a, vector a ; OQ = P , OP = xp; then T(OP.OQ) = k 2 , or xp 2 = -tf. ART. 37.] THE CIRCLE AND SPHERE. 75 But So. (xp-a) = Q; .. xSap = a*; k* hence p 2 = -~ Sap a is the equation of the locus of Q, which is therefore a circle, passing through 0. Ex. 5. Straight lines are drawn through a fixed point, to find the locus of the feet of perpendiculars on them from another fixed point. Let 0, A be the points, the lines being drawn through A. Let OA a, and let p = a + x(3 be the equation of one of the lines through A, 8 the perpendicular on it from 0. Then 8 = a + xfi, and S8 3 = SaS, because 8 is perpendicular to ft ; i.e. o*-SaS = 0, the equation of a circle whose diameter is OA. Ex. 6. A chord QR is drawn parallel to the diameter AB of a circle : P is any point in AB ; to prove that PQ* + PR* = PA* + PB*. Let CQ = P , CR = p, PC = a; then PQ* = - (vector PQ)' = -(a. + p}* = - (a 3 + 2Sap + p 3 ), PR* = - (a + p') 3 = - (a 2 + 2Sap' + p' 2 ) ; .-. PQ* + PR 2 = 2PC 2 + 2AC'-2 (Sap + Sap'). But S(p + p') (p - p) = and p p' = xa, because QR is parallel to AB \ . '. Sap + Sap' = 0, and PQ 2 + PR 2 = 2PC 2 + 2 AC 3 7G QUATERNIONS. [CHAP. V. Ex. 7. If three given circles be cut by any other circle, the chords of section will form a triangle, the loci of the angular points of which are three straight lines respectively perpendicular to the lines which join the centres of the given circles ; and these three lines meet in a point. Let A, B, C be the centres of the three given circles ; a, b, c their radii ; a, /?, y the vectors to A, B, C from the origin ', OA, OB, 00 respectively p, q, r ; D the centre of the cutting circle whose radius is R, OD = s, vector OD = 8, p the vector to a point of section of circle D with circle A ; then we shall have and .-. Now this is satisfied by the values of p to both points of sec- tion ; and being the equation of a straight line (32. 3) is the equation of the line joining the points of section of circle D with circle A call it line 1, and so of the others; then line 1 is 2S (8 - a) p = R 3 - a 2 - s* +p 2 , line 2 is 2S(& -p) P ' = R 2 -b*-s* + q 2 , line 3 is 2S(S - y) p" = H 2 -c 2 -s 2 + r 3 . If 1 and 2 intersect in P whose vector is p lt 1 and 3 in Q (p 2 ); 2 and 3 in R (p a ), we shall have by subtraction atP, therefore (32. 3) the lod of P, Q, JR are straight lines, perpen- dicular respectively to AH, AC, BC. Also at the point of intersection of the first and third of these lines, we have, by addition, which is satisfied by the second : hence the three loci meet in a point. ART. 37.] THE CIRCLE AND SPHERE. 77 Ex. 8. To find the equation of the cissoid. AQ is a chord in a circle whose diameter is AB, QN perpen- dicular to AB. AM is taken equal to BN, and MP is drawn perpendicular to A B to meet AQ in P ; the locus of P is the cissoid. Let vector AP = TT, AC = a, AM=ya, AQ = XTT; then y : 1 :: 2-y : x, by the construction ; Now is the equation of the circle ; 2SW M _ 7T* ' Also Tr a7r O.TT hence I 1 H -- 5- ) 5- = *i V TT / a and (7T 2 + 2^a7r) Sair = 2aV, is the equation required. Ex. 9. If ABCD is a parallelogram, and if a circle, be de- scribed passing through the point A, and cutting the sides AS, AC and the diagonal AD in the points F, G, H respectively ; then the rectangle AD . AH is equal to the sum of the rectangles AS . AF, andAG.AG. Let AF=xa, 78 QUATERNIONS. [CHAP. V. 6 the vector diameter of the circle ; then whence, since y = a + {3, zy* = xa* + yft 3 ; i.e. AD. AH = AB.AF+AC.AG. Ex. 10. What is represented by the equation If a, ft be not at right angles to one another, we can put a l + eft for a, and so choose e that Sa^ft = 0. We shall therefore consider a, ft as vectors at right angles to each other, and we may, on account of x, assume their tensors equal, and each a unit. a+xft a+xft Hence p = or, if I (a + xft)' ' 1+a?' p = - sin (a sin + ft cos 6), whence Tp (= r) = sin 0, a circle of which the diameter is a unit parallel to a and the origin a point in the circumference; and ft a tangent vector at the origin. Otherwise, Sap = or p 8 = Sap. AET. 38.] THE CIRCLE AND SPHERE. 79 Or, again, p" 1 = a + x{3 ; whence Sap' 1 = 1 , or VP (p~ l - a) = 0, where U stands for the versor of the quaternion ; all of these being, with the obvious condition S . a ftp = 0, varieties of the form of the equation of a circle, referred to a point in the circumference, the diameter through which is parallel to a. Draw any two radii p and p t , then we have S. U ' PiP Pl ' P 7 2 P wiu be rendered a unit if we take a unit PiP vector along each of the three vectors p 1? (p - p ; ), and p ; .-. s. U But and S. Up~ l U (p~ l - p- 1 ) Hence S. Up,U(p~ Pl } =-S/3Up. If p be constant whilst p l varies, the right-hand side of this equation is constant, and the equation shews that the angles in the same segment of a circle are equal to one another. Further, the form of the right-hand side of the equation, viz. SfiUp, shews that the angle in the segment is equal to the sup- plement of the angle between the chord (p) and the tangent (/?). 38. To draw a, tangent to a circle. 1. If we assume the first form of the equation, the centre being the origin, and assume also that the tangent is at right 80 QUATERNIONS. [CHAP. V. angles to the radius drawn to the point of contact ; we shall have, denoting by TT a vector to a point in the tangent, Sp (?r - p) = 0, for TT p is along the tangent ; . \ Sirp a 2 is the equation required. 2. Without assuming the property of the tangent, we may obtain it as follows. Let p be a point in the circle near to P ; then from the equation ; But p' 4- p is the vector which bisects the angle between the vectors to the points of section, and p p is a vector along the secant. Now the equation shews (22. 7) that the former of these lines is perpendicular to the latter. As the points of section approach one another, the tangent approaches the secant, and the bisecting line approaches the radius to the point of contact : therefore the radius to the point of contact is perpendicular to the tangent. 39. From a point without a circle two tangents are drawn to the circle, to find the equation of the chord of contact. Let /3 be the vector to the given point, <? ~a / / \\ I ~ ^ / I \ n the equation of a tangent; then since it passes through the given point Now this equation is satisfied for both points of contact, and since it is the equation of a straight line (32. 3) it must be satis- fied for every point in the straight line which passes through those points : it is therefore the equation of the chord of contact. To ART. 40.] THE CIRCLE AND SPHERE. 81 avoid the appearance of limiting p to a point in the circle, we may write a- -in place of p ; and the equation of the chord of contact becomes Spa- = - a 3 . EXAMPLES. 40. Ex. 1. If chords be drawn through a given point, and tangents be drawn at the points of section, the corresponding pairs of tangents will intersect in a straight line. Let y be the vector to the given point G, the centre C being the origin ; ft the vector to 0, the point of intersection of two tangents at the extremities of a chord through G ; then the equa- tion of the chord of contact is (39) S/3<r=-a s , and as the chord passes through G we have which, since y is a constant vector, is the equation of a straight line, the locus of ft. COR. 1. The straight line is at right angles to CG (32. 3). COR. 2. The converse is obviously true, that if through points in a straight line pairs of tangents be drawn to a circle, the chords of contact all pass through the same point. Ex. 2. Any chord drawn from the point of intersection of two tangents, is cut harmonically by the circle and the chord of contact. Let radius = a, 0(7 -c, OR=p, OS=q, vector OC=a, unit vector OR = p ; then is the equation of the circle ; i.e. p 1 + 2pSap + c" - a 3 0, T. Q. QUATERNIONS. [CHAP. v. a quadratic equation which gives the two values of p, viz. OR and OT; JL _L. 2Sa P ' 777? O7 r ~ /* // 2 " L/.Zli \s J- C tw Saqp = SaON; hence 2__2 0^~g 2Sap _L ! ~0/i + OT' Ex. 3. 7/" tangents be drawn at the angular points of a triangle inscribed in a circle, the intersections of these tangents with the opposite sides of the triangle lie in a straight line. Let radius = a, OA = a, OE = fi, 00 = 7, then ART. 40.] THE CIRCLE AND SPHERE. But a is perpendicular to AP ; 83 a' + Sap Say -Sap' and Say - Sap oap opy Sim ii arly , ^ Say Hence (Say - Saft) OP + (Sap - Spy) OQ + (Spy -Say) OB = 0, whilst (Say - Sap) + (Sap - Spy) + (Spy - Say) = 0. Consequently (Art. 13) P, Q, R are in the same straight line. COR. PQ : PR :: Spy-Say :: Spy-S a p :: cos 2.C cos 2 J. : cos 2(7 cos 2A :: sin C sin (B - A) : Ex. 4. A faced circle is cut by a number of circles, all of which pass through two given points ; to prove that the lines of section of the fixed circle with each circle of the series all pass through a point whose distances from the two given points are proportional to the squares of the tangents drawn from those points to the fixed circle. Let be the centre of the fixed circle whose radius is a, A, B the given points, vectors a, p, the origin being ; OA = b, OB c;C the centre of a circle which passes through A and B, radius r 00 = p, TT the vector to any point in the circumference of this circle; then the equation of the circle is (TT p) 2 = r 2 ; 62 84 QUATERNIONS. [CHAP. V. hence for the four points A, JB, P, Q, we have a 2 - 2Sap + p 2 = - r\ From which it follows that S(OP-OQ) P = ....................... (1), -b* + c 2 = a:-p z = 2S(a-p)p ................ (2), 2S(OP-a)p=OP 2 -a 2 = -a i +b 2 ............... (3). Let QP, AB intersect in R, OR = a- ; then = S.OP P \y(l), and Sa-p = S {a + y (a - /?)} p = 2S(OP-a)p =-sf + V\ty(S), i.e. y is independent of p and r ; or R is the same point for every circle : (c'-tt')a-(6'-a')0 also OR = - - 1-3 ^ - H- , c o and RA : RB :: a- OR : ft- OR :: V-a* : c*-a* :: AT 2 : BU*. 41. The Sphere. 1. It is clear that there is nothing in the demonstration of Art. 36 which limits the conclusions to one plane ; it follows that the equations there obtained are also equations of a sphere. 2. Further if we assume that the tangent plane to a sphere is perpendicular to the radius to the point of contact, the con- clusion in Art. 38 is applicable also. ART. 42.] THE CIRCLE AND SPHERE. 85 The equation of the tangent plane to the sphere is therefore 3. Lastly, the results of Art. 39 are also applicable if we substitute any number of tangent planes passing through a given point for two tangent lines ; the equation of the plane which passes through the points of contact is therefore S/3<r=-a*. This plane is the polar plane to the point through which the tangent planes pass. COR. Since the polar plane is perpendicular to the line which joins the centre with the point through which the tangent planes pass, the perpendicular CD to it from the centre is along this line and has therefore the same unit vector with it. The equa- tion above gives in this case .-. CO. CD = a 2 (19). EXAMPLES. 42. Ex- ! Every section of a sphere made by a plane is a circle. Let p 2 = a 8 be the equation of the sphere, a the vector per- pendicular from the centre on the cutting plane ; c the correspond- ing line. Let p = a + TT ; then the equation becomes But Sair = ; .-. 7r 2 = -(a 2 -c 2 ) is the equation of the section, which is therefore a circle, the square of whose radius is a 2 c 2 . Ex. 2. To find the curve of intersection of two spheres. Let the equations be p 2 -2Sa.' P =C'; 86 QUATERNIONS. [CHAP. V. .'. 2S(a'-a)p=C-C', a plane perpendicular to the line of which the vector is a' a, i.e. to the line which joins the centres of the two spheres. Hence, by Ex. 1, the curve of intersection is a circle. Ex. 3. To find the locus of the feet of perpendiculars from the origin on planes which pass through a given point. Let a be the vector to the point, 8 perpendicular on a plane through it ; then is the equation of that plane ; therefore for the foot of the per- pendicular S(S 2 -aS)=0; or S 2 -SaS = is true for the foot of every perpendicular and is therefore the equation of the surface required. Hence it is a sphere whose diameter is the line joining the origin with the given point. Ex. 4. Perpendiculars are drawn from a point on the surface of a sphere to all tangent planes, to find the locus of their extremi- ties. Let a be the vector to the given point, Sirp = a" the equation of a tangent plane. Since the perpendicular is parallel to p, its vector is TT = a + xp ', because both p and a are vector radii. But Sirp = a* gives with xp = TT - a, STT (IT a) = a*x, (** - Sair)* = a*x s = a 2 x a*x' = -a a (7r-a) f . ART. 42.] THE CIRCLE AND SPHERE. 87 Ex. 5. If the points from which tangent planes are drawn to a sphere lie always in a straigM line, prove that the planes of sec- tion all pass through a given point. Let CE be perpendicular to the line in which the point ft lies (41), see fig. of Art. 39, CE=c, vector CE=8; then SpS = -c> is the equation of the line. But Sp<r = -a* is the plane of contact, which is therefore satisfied by i. e. the planes all pass through a point G in CE, such that CG = - a CE, or CE.CG=a\ Ex. 6. If three spheres intersect one another, their three planes of intersection all pass through the same straight line. Let a, /?, y be the vectors to the centres of the three spheres, p'-2Sap=a, their three equations ; .-. 2S (a y) p = c - a, are the equations of the three planes of intersection. Now the line of intersection of the first and second of these planes is obtained by taking p so as to satisfy both equations, and therefore their difference 88 QUATERNIONS. [CHAP. V. which, being the third equation, proves that the same value of p satisfies it also. The three planes consequently all pass through the same straight line. Ex. 7. To find tlie locus of a point, the sum of the squares of whose distances from a number of given points has a given value. Let p denote the sought point ; a, /?,... the given ones ; then If there be n given points ; this is or A>_! a y = (^y_!(2.a 2 +(7). \ r n } \n J n^ This is the equation of a sphere, the vector to whose centre is -2 (a), n i. e. the centre of inertia of the n points taken as equal. Transpose the origin to this point, then (36) and /' = - {* (a f ) + 0}- Hence, that there may be a real locus, C must be positive and not less than the sum of the squares of the distances of the given system of points from their centre of inertia. If C have its least value, we have of course *> = 0, the sphere having shrunk to a point. ADDITIONAL EXAMPLES TO CHAP. V. 1. If two circles cub one another, and from one of the points of section diameters be drawn to both circles, their other extre- mities and the other point of section will be in a straight line. EX. 2.] ADDITIONAL EXAMPLES. 89 2. If a chord be drawn parallel to the diameter of a circle, the radii to the points where it meets the circle make equal angles with the diameter. 3. The locus of a point from which two unequal circles sub- tend equal angles is a circle. 4. A line moves so that the sum of the perpendiculars on it from two given points in its plane is constant. Shew that the locus of the middle point between the feet of the perpendiculars is a circle. 5. If 0, 0' be the centres of two circles, the circumference of the latter of which passes through ; then the point of inter- section A of the circles being joined with 0' and produced to meet the circles in G, D, we shall have 6. If two circles touch one another in 0, and two common chords be drawn through at right angles to one another, the sum of their squares is equal to the square of the sum of the diameters of the circles. 7. A, , G are three points in the circumference of a circle; prove that if tangents at E and G meet in D, those at C and A in E, and those at A and B in F; then AD, BE, CF will meet in a point. 8. If A, B, G are three points in the circumference of a circle, prove that V (AB . BC . CA) is a vector parallel to the tan- gent at A. 9. A straight line is drawn from a given point to a point P on a given sphere : a point Q is taken in OP so that OP.OQ^k 3 . Prove that the locus of Q is a sphere. 10. A point moves so that the ratio of its distances from two given points is constant. Prove that its locus is either a plane or a sphere. 90 QUATERNIONS. [CHAP. V. 11. A point moves so that the sum of the squares of its distances from a number of given points is constant. Prove that its locus is a sphere. 12. A sphere touches each of two given straight lines which do not meet ; find the locus of its centre. CHAPTER VI. THE ELLIPSE. 43. ! I* 1 "we define a conic section as "the locus of a point which moves so that its distance from a fixed point bears a con- stant ratio to its distance from a fixed straight line " (Todhunter, Art. 123), we shall find the equation to be (Ex. 5, Art. 35) ay = e 8 (a 2 -Sap) 2 (1), where SP = ePQ, vector SD = a, SP = p. When f is less than 1, the curve is the ellipse, a few of whose properties we are about to exhibit. 2. SA, SA' are multiples of a : call one of them xa : then, by equation (1), putting xa for p, we get 92 QUATERNIONS. [CHAP. VI. +e .-. AA'=^ 9 SD, J. ~~ 6 the major axis of the ellipse, which we shall as usual abbreviate by 2a. If C be the centre of the ellipse J.-e 1e* = ae, and if vector CS be designated by a, CP by p, we have I a, and p' p + a' ; L whence, by substituting in (1), the equation assumes the form ay 2 +(&*>')' = - a 4 (1-e 2 ); which we may now write, CS being a and CP p, -a 4 (l-e 2 ) .................... (2). 3. This equation might have been obtained at once by re- ferring the ellipse to the two foci, as Newton does in the Prin- cipia, Book i. Prop. 1 1 ; the definition then becomes or in vectors, if i. e. J- (p + a) i + J-(p- a)" = 2a ; hence, squaring, a J (p a) 2 = a 3 + Sap ; ART. 45.] THE ELLIPSE. 93 If now we write <t>p for --- , where <4p is a vector a (1 - e) which coincides with p only in the cases in which either a coin- cides with p or when Sap = 0, i. e in the cases of the principal axes ; the equation of the ellipse becomes 1 ............................... (3). The same equation is, of course, applicable to the hyperbola, e being greater than 1. 44. The following properties of <f>p will be very frequently employed. The reader is requested to bear them constantly in mind. 1. < (p + <r) = <f>p + (fur. = X(f>p. a 2 S<rp 4- SarrSap a* (1-0 They need no other demonstration than what results from simple inspection of the value of <p a'p + aSap ~ *(i-o ' 45. To find the equation of the tangent to the ellipse. The tangent is defined to be the limit to which the secant approaches as the points of section approach each other. Let CP = p, CQ = p f , then vector PQ = CQ - GP = p - p = ft say ; j8 is therefore a vector along the secant. Now Sp'<t>p =S( P + P)<f>(p + P) 4>P) (44.1) QUATERNIONS. [CHAP. VI. But Sp'fo' = 1 = Sp<f>p 5 or (44. 3) 2Sp<f>p + SP<I>P = 0. Now P<f>p involves the first power of P whilst P^p involves the second, and the definition requires that the limit of the sum of the two as P gets smaller and smaller should be the first only, even if that should be zero : i. e. when P is along the tangent, we must have = 0. [We might also have written the equation in the form lff.0(*H-3tf),-* Thus, however small the tensor of P may be, is always perpendicular to /?. Whence, finally, /?&> = 0.] Let then T be any point in the tangent, vector CT = IT, then it = p + xp, and Sp<f>p = gives . '. Sir<f>p = Sp(j>p = 1 is the equation of the tangent. COR. 1 <}>p is a vector along the perpendicular to the tangent (32. 3), that is, <f>p is a normal vector, or parallel to a normal vector at the point p. COE. 2. The equation of the tangent may also be written (44. 3) Spfir = 1. 46. We may now exhibit the corresponding equations in terms of the Cartesian co-ordinates, as some of the results are best known in that form. ART. 46.] THE ELLIPSE. 95 Let CM=x, MP y as usual; then, retaining the notation of Art. 31 with i, j as unit vectors parallel and perpendicular respectively to CA, vector CM- xi, HP = yj, OS - aei ; .-. p = xi + yj, a*p + aSap * p= "a*(l-O a* (1 e*) xi + cfyj , 30 'tf+ where 6 2 -a s l-e s ; and . L 4.2-ssl ' a 3 b'~ is the Cartesian interpretation of Sp<f>p = 1. Again, if x', y be the co-ordinates of T a point in the tangent, and S*<t> P = -S (x'i + y'j] + is the equation of the tangent. 96 QUATERNIONS. [CHAP. VI. 47. The values of p and <j>p exhibited in the last Article, viz. enable us to write a We shall have , t , . = 9<PP = If, further, we write \}/p for + we shall have l p = (aiSip 4- bjSjp), &c. P = 'TVp (5). It is evident that the properties of <f>p (Art. 44) are possessed by all these functions. Now gives Sp\j/ (if/p) --1. But since Sptyo- = this becomes fyp^P = or Ttyp = 1 ; ART. 48.] THE ELLIPSE. 97 which shews 1. that if/p is a unit vector; 2. that the equation, of the ellipse may be expressed in the form of the equation of a circle, the vector which represents the radius being itself of vari- able length, deformed by the function \j/. Lastly, Sa<j>(3 = gives Scupp = Sijnolrp = ; therefore if/a, \f/(3 are vectors at right angles to one another. 48t To find the locus of the middle points of parallel chords. Let all the chords be parallel to the vector (3 ; TT the vector to the middle point of one of them whose vector length is 2x(3 ; then TT + xfi, ir xft are vectors to points in the ellipse ; multiplying out, observing that (44. 1), <f> ( we get by subtracting, = 0, or, (Art. 44. 3), 2Sir<t>P = ; .-. Sv<l>p = 0, i. e. the locus required is a straight line perpendicular to </>/?. Now (f>fi is the vector perpendicular to the tangent at the extremity of the diameter ft (Art. 45. Cor. 1). Therefore the locus of the middle points of parallel chords is a diameter parallel to the tangent at the extremity of the diameter to which the chords are parallel. COB. If a be the diameter which bisects all chords parallel to (3 - } since Sa<j>P = 0, T. Q. 7 98 QUATERNIONS. [CHAP. VI. we have (Art. 44. 3), Sft<j>a = 0, which is the equation to the straight line that bisects all chords parallel to a. Moreover ft is parallel to the tangent at the ex- tremity of a, for it is perpendicular to the normal </>a. Hence the properties of a with respect to ft are convertible with those of ft with respect to a : and the diameters which satisfy the equation Sa<t>ft = 0, are said to be conjugate to one another. 49. O ur object being simply to illustrate the process, we shall set down in this Article a few of the properties of conjugate diameters without attempting to classify or complete them. 1. If CP, CD are the conjugate semi- diameters a, ft - } and if DC be produced to meet the ellipse again in E, and PD, PE be joined ; vector DP = a ft, vector EP = a + ft. Now = Sa<t>a-Sft<}>ft-Sa<j>ft+Sft<l>a(U. 1) = o, because Safa, Sft^ft, each equals 1. Therefore a + ft, a ft are parallel to conjugate diameters. (Art. 48. Cor.) This is the property of /Supplemental Chords. 2. Let two tangents meet in T, CI'=-n; and let the chord of contact be parallel to ft. If for the present purpose we denote CN by a, we have (a + a, ft) = 1, for the two points of contact. ART. 49.] THE ELLIPSE. 99 Subtracting and applying (44. 1), &*<}>($ = : hence TT and ft Le. CT, QR are conjugate. 3. The equation of the chord of contact is S<T^TT = 1. For Spfar = 1 (45. Cor. 2) is satisfied by the values of p at Q and at B, and since Sp<f>ir = 1 or S<r<}*ie = 1 is the equation of a straight line, ir being a constant vector (32. 3) it is the line QR. 4. If QR pass through a fixed point JZ } the locus of T is a straight line. Let <r be the vector to the point E, then Sa-tfrir = 1 ; .'. /S f 7r^r = l, or the locus of T is a straight line perpendicular to <fxr, i.e. parallel to the tangent at the point where CE meets the ellipse. (45. Cor. 1.) The converse is of course true. 5. Let us now take CP = a, C = p, CN=xa, NQ = y p, CT=za' } 72 100 QUATERNIONS. [CHAP. VI. then the equation of the tangent becomes Sza<j> (xa + yP) = I ; i.e. xzSafya = 1 ; .: xz = 1, or xa.za.-a* ; geometrically CN.CT= CP S . 6. The equation of the ellipse gives S(xa + yp) <f> (xa + yp) = 1, or x*Sa<}>a i.e. or, since CN is xa, CP-a, &c., \CPJ + \CD. the equation of the ellipse referred to conjugate diameters. 7. a = if/~ l ifra = - (aiSi\f/a + bjSjij/a) . '. Vap db Vij (Siij/aSj\}/P Si\ If now we call k the unit vector perpendicular to the plane of the ellipse, we get Vij = k. And, observing that ij/a, \f/p are unit vectors at right angles ; if the angle between i and tya be 0, that between i and \j/(3 will be - + 6, &c. &c., we shall have (21. 3) Sifya = COS 6, = sin0, tya = sin 0, = cos 0. ida = cos 8 6 + sin* = 1. ART. 50.] THE ELLIPSE. 101 Consequently Fa/3 = able ; or all parallelograms circumscribing an ellipse are equal. 50. EXAMPLES. Ex. 1. To find the length of the perpendicular from the centre on the tangent. Let CY the perpendicular, which (Art. 45. Cor. 1) is a vector along </>p, be x<f>p ; then since T is a point in the tangent, 1 gives /Sx<f>p<f>p = 1, or x (<j>p) a = 1 ; and (46). Ex. 2. The product of the perpendiculars from the foci on tJie tangent is equal to the square of the semi-axis minor. We have SY the vector perpendicular = x<f>p, and as Y is a point in the tangent, and x (<p) 2 = 1 Saxftp, 9P Similarly, E Z=T l -^-', <PP 102 QUATERNIONS. [CHAP. VI. Now (43. 2) V - - S 2 ap -a 4 (I- e 2 ), a?p + aSap 4 Cf% r - /3 ap Ex. 3. ^Ae perpendicular from the focus on the tangent in- tersects the tangent in the circumference of the circle described about the axis major. Retaining the notation of the last example, we have CY=a + x<j>p $p(l- Sa<f>p) ~~ 2 ., = aV a 2 (1 e z ) (last example) and the line CY=a. Ex. 4. To ^c? i/ie locus of T when the perpendicular from the centre on the chord of contact is constant. If CT be TT, the equation of QR, the chord of contact, is 7r=l (Art. 49. 3), and the perpendicular (Ex. 1) is T ; ART. 50.] THE ELLIPSE. 103 .-, (**)= -c", Or /S<f>7T . <f>TT = C 2 , or &r^r = -c* (Art. 44. 3); x 2 y* r + l? = ^ an ellipse. Ex. 5. FQ, TR are two tangents to an ellipse, and CQ', CR' are drawn to the, ellipse parallel respectively to TQ, TR ; prove that Q'R' is parallel to QR. Let CQ=p, CR = P ', CT=a, then Sp<f>a = 1, Now since CQ' ,is parallel to TQ, CQ'=xTQ = x(p- Similarly CR' = y (p - a), and gives y?S (p a) <f> (p a) = 1, i.e. x*(Sa<j>a-l) = I, and y 2 (Sa^a - 1) = 1; . . y = x, and = xQR; hence Q'R' is parallel to QR. COR. Q'R 2 : QR 3 :: x 2 : 1 :: 1 : Safa- where aj, y are the co-ordinates of T. 104 QUATERNIONS. [CHAP. VI. Ex. 6. If a parallelogram be inscribed in an ellipse, its sides are parallel to conjugate diameters. Let PQRS be the parallelogram. then CQ = p + a, CR = p+a; .'. Sp<f>p=l, wherefore %Sp<$>a + Sa<^a = 0. Similarly 2Sp'<j>a + Sa<j>a = ; .'. S(p' p) <a = 0, by subtraction, or Sp<f>a = 0, and (48. Cor.) /?, a are parallel to conjugate diameters. ADDITIONAL EXAMPLES TO CHAP. VI. 1. Shew that the locus of the points of bisection of chords to an ellipse, all of which pass through a given point, is an ellipse. 2. The locus of the middle points of all straight lines of con- stant length terminated by two fixed straight lines, is an ellipse whose centre bisects the shortest distance between the fixed lines; and whose axes are equally inclined to them. 3. If chords to an ellipse intersect one another in a given point, the rectangles by their segments are to one another as the squares of semi-diameters parallel to them. 4. If PGP', BCD' are conjugate diameters, then PD, PD' are proportional to the diameters parallel to them. 5. If Q be a point in the focal distance SP of an ellipse, such that SQ is to SP in a constant ratio, the locus of Q is a similar ellipse. EX. 6.] THE ELLIPSE. 105 6. Diameters which coincide -with the diagonals of the paral- lelogram on the axes are equal and conjugate. 7. Also diameters which coincide with the diagonals of any parallelogram formed by tangents at the extremities of conjugate diameters are conjugate. 8. The angular points of these parallelograms lie on an ellipse similar to the given ellipse and of twice its area. 9. If from the extremities of the axes of an ellipse four pa- rallel lines be drawn, the points in which they cut the curve are the extremities of conjugate diameters. 10. If from the extremity of each of two semi-diameters ordinates be drawn to the other, the two triangles so formed will be equal in area. 11. Also if tangents be drawn from the extremity of each to meet the other produced, the two triangles so formed will be equal in area. 12. If on the semi-axes a parallelogram be described, and about it an ellipse similar and similarly situated to the given ellipse be constructed, any chord PQR of the larger ellipse, drawn from the further extremity of the diameter CD of the smaller ellipse, is bisected by the smaller ellipse at Q. 13. If TP, TQ be tangents to an ellipse, and PCF be the diameter through P, then PQ is parallel to CT. CHAPTER VII. THE PARABOLA AND HYPERBOLA. 51. As already stated, most of the properties of the hyperbola are the same as the corresponding properties of the ellipse, and proved by the same process, e being greater than 1. There are, however, some properties both of it and of the parabola which may be conveniently developed by a process more analogous to that of the Cartesian geometry. This process we shall develope presently. In the meantime we proceed to give a brief outline of the application to the parabola of the method employed in the preceding Chapter for the ellipse. 52. If S be the focus of a parabola, DQ the directrix, we have SP = PQ, SA=AD = a. If SP = p, SD = a, we have (Ex. 5, Art. 35) a a p a = (a 3 -SapY (1). p a" 1 /Sap If <>=' (2), to which the properties of cj>p in Art. 44 evidently apply, the equation becomes Sp (<f>p + 2a- J ) = 1 If p r be another point in the parabola, p' p = /?, the which /3 approaches is a vector along the tangent ; so xf} = ir-p, TT is the vector to a point in the tangent ; this ..(3). limit to that if gives ART. 52.] THE PARABOLA AND HYPERBOLA. 107 hence the equation of the tangent becomes Sir(<l>p+a- l )+Sa- l p=l ................... (5). From (2) it is evident that so that <f>p is a vector perpendicular to the axis. From the same equation a From (4) the normal vector is tp+a- 1 ............................ (8); therefore the equation of the normal is <r = p + x (<t>p + a" 1 ) ....................... (9). Equation (2) when exhibited as a 2 (f)p = p a~ l /Sap, reads by (6), 'vector along NP = SP - vector along AN', which requires that a*<j>p ............................ (10), i.e. =cuSa-*p ......................... (11). For the subtangent AT, put xa for TT in (5), and there results by (6) X + Sa l p = ~i, whence \ x ~ 9J a = 9 a "~ a ^ a ~V > i. e. vector A T = - vector AN (by 11); 108 QUATERNIONS. [CHAP. VII. and ST=xa gives S2" = (a-aSar l p) a _(a'-Sap) s .-. line ST=SP, whence also the tangent bisects the angle SPQ ; and SQ is per- pendicular to and bisected by the tangent. From (8) y ($p + a~ l ) = PG = PN+ NG = - a a <f>p + zo. (by 10) ; .-. y = -o*, y = za s , i. za = a ) Le. NG = -SD, or linQNG = SD, whence the subnormal is constant. And vector GP --y (<f>p + a~ l ) - a? (<j>p + a" ') ; .-. vector SQ = SD+DQ and SQGP is a rhombus. Lastly, = a + a.*(j)p or (10) A Y is parallel to, and equal to half of NP. ART. 54.] THE PARABOLA AND HYPERBOLA. 109 53. If n ow we substitute Cartesian co-ordinates, making p = xi + yj, a = -2ai; we shall have C* 1 *^ ^ a p =~2a' a~ 1 Sap = xi, ^ = ~^ ; and equation (3) becomes J^_*_l 4a 8 a or y* = a(a + x) = 4ax' if x' = AN. The locus of the middle points of parallel chords is thus found. Let the chords be parallel to (3, TT the vector of the middle point of one of the chords, then and X which, since the term involving x must disappear, gives a straight line perpendicular to </?, i. e. (6) parallel to the axis. This equation may be written tf (<*+ a" 1 ) = 0, which shews (8) that the chords are perpendicular to the normal vector at the point where P = TT, i.e. at the point where the locus of the chords meets the curve : in other words, the chords are parallel to the tangent at the extremity of the diameter which bisects them. 54. EXAMPLES. Ex. 1. If two chords be drawn always parallel to given lines, and cut one another at points either within or urithout the parabola, 110 QUATERNIONS. [CHAP. VII. the ratio of the rectangles of their segments is always the same whatever be their point of section. Let POp, QOq be the chords drawn through 0, and always parallel respectively to /3 and y, which we will suppose to be unit vectors. Let 8 be the vector to 0, then p-S + x/3 gives from equation (3) the product of the two values of x being a constant ratio whatever be 0. Con. Let 6, & be the angles in which P and y cut the axis ; then since /3, y are unit vectors, if p be a vector to the parabola, drawn from S parallel to POp, which we may now call SP ; P = n(3, <fr> = ^(w) = n^8(44. 2), will ive . NP in which case <pp is j- ; a : Sy<f>y :: sintf-^ : sintf'-^- :: sin 2 : sin 2 6'; and, OP . Op : OQ . Oq :: . a/i : . a/v . sin sin o Ex. 2. jFwc^ f/ie locus of the point which divides a system of parallel chords into segments whose product is constant. AET. 54.] THE PAEABOLA AND HYPERBOLA. Ill By the last example, the equation of the locus is a parabola similar to the given parabola. Ex. 3. The perpendicular from A on tJte tangent, and the line PQ are produced to meet in R : find the locus of R. By Art. 52. 8, AR = x (<f> P + a' 1 ), and PR = ya ; ~ Operate by S<j)p, and x (<p) 2 = Sp<j>p (52.7); and TT = <T + a 2 (<j>p + a" 1 ) O = -H- + a*^P is the equation required ; (O V TT ^- j a 0, it is that of a straight line perpendi- cular to the axis, at the distance 3a from 8. Ex. 4. Jb ^/md tf/^e focws of the intersection with t/te tangent of the perpendicular on it from the vertex. If TT be the vector perpendicular on the tangent from A, we have by (52. 8) TT = x (0p + a' 1 ) .......................... (1), and the equation of the tangent gives, putting TT + ^ in place a of TT in (52. 5), and multiplying by 2, - 2/ffir^p + 2/S r a-'ir + a^o^p = 1 ................. (2), we have also Sp (<j> P + 2a- J ) = 1 ..... ................ (3). 112 QUATERNIONS. [CHAP. VII. From these three equations we have to eliminate x and p. Equation (1) gives SO.TT - x, which gives x, and . Str(f)p = x (<f>p) 2 , which substituted in (2) gives Also, substituting (52. 7) a* (<j>p) 3 for Sptfrp, equation (3) gives therefore by subtraction (2x - a 2 i. e. (2Sa.Tr - a 2 ) (<f>p)* + 2Sa- l ir = 0, which from (1) becomes, multiplying by S*air, (2Sair - a) 2 (n - aT l Sa.Tr} 2 + 2S 2 a7nS f cr 1 7r = 0. This equation at once reduces to 27r e #x7r - TT-V + S*cnr = 0, an equation which, when 4a is written in place of a, becomes identical with that obtained in Art. 37, Ex. 8. The locus is therefore a cissoid, the diameter of the generating circle being AD. 55. It will probably have suggested itself to the reader, that there exists a large class of problems to which the processes we have illustrated are scarcely if at all applicable. Hence there may have arisen a contrast between the Cartesian Geometry and Quaternions unfavourable to the latter. To remove this un- favourable impression, all that is required in a reader familiar with the older Geometry is a little experience in combining the logic of the new analysis with the forms of the old. He will then see how simple and direct are the arguments which he can bring to bear on any individual problem, and consequently how little the memory is taxed. ART. 55.] THE PARABOLA AND HYPERBOLA. 113 We propose in this Article to put the reader in the track of employing his old forms in conjunction with quaternion reasonings. We shall work several examples on the parabola and the hyperbola. Having applied quaternions pretty fully to the ellipse in what has preceded, we will limit ourselves to a single example in this case. 1. The Parabola. If the unit vector along any diameter of the parabola be a, and the unit vector parallel to the tangent at its extremity be ft; we may write the equation of the parabola under the form For the particular case in which the diameter in question is the axis, and the tangent at its extremity parallel to the directrix where a is A S (Art. 52). This is the most convenient form when the focus is referred to. In other cases a somewhat simpler form may be obtained by supposing a, or if necessary both a and /3 of equation (1) to be other than unit vectors. The equation may then be written under the form P = 2* + *P (3). To find the equation of the tangent, we have T. Q. QUATERNIONS. [CHAP. VII. Now p p is a vector along the secant; and its limit is a rector along the tangent : hence any vector along the tangent is a multiple of to. + /? ; and the equation of the tangent may be written (4). EXAMPLES. Ex. 1. If AP, AQ be chords drawn at rigid angles to one another from A ; PM, Q<& perpendiculars on tJie axis, then the latus rectum is a mean proportioned between AM and AN ; or between PM and QN. If PJf=y, QK=y, , Now S(AP.AQ)=0(22. 7); or yy= therefore also aai - Ex. 2. If the rectangle of ichich AP, AQ are the fides be completed, the further angle witt trace out a parabola similar to the given parabola, tfe distance between the tico vertices being equal to twice the latus rectum. Ex. 3. The circle described on a focal chovd as diameter touch fs the directrix; and the circle described on any other chord dots not reach tfte directrix. ART. 55.] THE PARABOLA AND HYPERBOLA. 115 Let PQ be any chord, centre 0, The equation of the circle with centre 0, radius OP, is AQ-AP\* / ( p - 2 or p-S(AP + At the points in which this circle meets the directrix p = aa -f s/3 ; or This equation is possible only when yy+4a*=0; i. e. when the chord is a focal chord. I/ J. T/' In this case the two values of z are equal, each being ( - ; A and the directrix is a tangent to the circle. Ex. 4. Two parabolas have a common focus and axis ; their vertices are turned in opposite directions. A focal chord cuts them in PQ, P'Q', so that PP'SQQ' are in order. Prove (1) that SP.SP = SQ.SQ'; (2) that SP : SQ' ia a constant ratio; and (3) that the tangents at P, f are at right angles to one another. The equations of the parabolas are V 3 + the focus being the origin. 82 116 QUATERNIONS. [CHAP. VII. Now since p, p are in the same straight Hue when the common chord is the focal chord, we have p'=pp; y'=py, ' (yy' - 4oa') (ay + ay') = 0. Taking the former factor, we must have y, y' on the same side of the axis with a constant product; therefore The second factor gives SP : SQ' a constant ratio a : a'. Lastly, by Equation (4), the tangent vectors at P and P' are parallel to therefore the tangents are at right angles to one another. Ex. 5. If a triangle be inscribed in a parabola, the three points in which ilie sides are met by the tangents at the angles lie in a straight line. Let OPQ be the triangle. Take as the origin, then t* r=2<>- t r * ' ' ' ART. 55.] THE PARABOLA AND HYPERBOLA. 117 are the vectors OP, OQ, and the equations of the tangents at P and Q. If QO meet in A the tangent at P, t 2 9 y> t + x- t'y, t 3 and Similarly if the tangent at Q meets PO in B, If the tangent at meets PQ in (7, OC=OP + z(PQ) = OP + z(OQ-OP) t 3 (t" t 2 = -o + $ + *| 2 a + But OC = v(3; 2 T ~2~ t + z(^ t) v, ft' QUATERNIONS. [CHAP. VII. Now 2t-t' It'-t f-t" and also --- j- -g- 0; therefore (Art. 13) A, B, C are in a straight line. 2. The ellipse. If a, ft are unit vectors along the axes, the equation of the ellipse may be written b* where y* = -5 (a 2 - a; 8 ) = m (a 2 - # 2 ) ; Gb and the equation of the tangent will be readily seen to be ir = xa + y(3 + X (ya - mxfi). A single example will suffice. Ex. If tangents be drawn at three points P, Q, R of an ellipse intersecting in R', Q', P, prove tJiat, PR'. QF. RQ' - PQ'. QR'. RP. If x, y; x', y ; x", y" are respectively the co-ordinates of P, Q, R', we shall have CR' xa + y($ + X (ya. - mxft) - x'a + y'fi + X ' (y'a - mx'fi) ; y mXx = y' mX'x' y . \ mZ (x'y - y'x) = mx' a + y' 2 - mxx' - yy' = b a mxx' yy' . Hence mX ' (xy - x'y) -b a - mxx' - yy' = -mX(xy'-x'y); .-. X=-X', Y = -Y' for<?', Z = -Z' for/", and ART. 55.] THE PARABOLA AND HYPERBOLA. 119 Now X PR' . = hence the proposition. 3. The hyperbola. If a, 8 are unit vectors parallel to the asymptotes CX, CY, the equation of the hyperbola may be written since = xa + - B. x ,= a ^-=G. If a, /3 be not both units we may write the equation under the simpler foi*m P = a + .............................. (1). To find the equatioa of the tangent, we have as usual a vector parallel to the secant and a vector parallel to the tangent will be 120 QUATERNIONS. [CHAP. VII. Hence the equation of the tangent is TT = to. + - + x ( to. - '-, t COR. It is evident that are conjugate semi-diameters. EXAMPLES. Ex. 1. One diagonal of a parallelogram tcJtose sides are tJie co-ordinates being the radius vector, the other diagonal is parallel to the tangent. We have CN = ta, tfQ = % , t 7> and the other diagonal is which, equation (2), is parallel to the tangent at Q. Ex. 2. Any diameter CP bisects all the chords which are parallel to the tangent at P. Let CP be to. + - , t then the tangent at P is parallel to .-f; But as Q is a point in the hyperbola, this equation must have the form ART. 55.] THE PARAIOLA AND HYPERBOLA. 121 and X*-Y a =l, an equation which gives two equal values of Y with opposite signs, for every value of X. Hence all chords are bisected. COR. X'-Y a = lia f2KJ/fi? \VP) \CD CD being ta-@ = PO. t This is the ordinary equation of the hyperbola referred to conjugate diameters. Ex. 3. If TQ, T'Q' be two tangents to tJte hyperbola intersect- ing in R and terminated at T, T', Q, Q' by the asymptotes; then (1) TQ' is parallel to T'Q; (2) area of triangle TRT' = area of triangle QRQ', and (3) CR bisects TQ' and T'Q. The equation of the tangent gives fiW Of. I/ J. Ml (the coefficient of /3 being 0), t CT = 2t' * v therefore Q'T is parallel to QT'. 122 QUATERNIONS. [CHAP. VII. Again, CR = CQ+QR = CQ + IB f B\ Also CR = ~ + x'2(at'- p ,}; t \ t .: xt = x't', 1 x 1 tf t' ~t+t" ,_ t ~ t + 1'' and xx' = ( 1 - a;) (1 a;'), and the triangles TRT', QRQ' are equal Lastly, C - t t+t\ t or CR is in the direction of the diagonal of the parallelogram of which the sides are CT, CQ' ; and therefore CR bisects TQ' and T'Q. Ex. 4. If through Q, P, Q' parallels be drawn to CX meeting CY in E, F, G ; CE, CF, CG are in continued proportion. t = GV+VQ ART. 55.] THE PARABOLA AND HYPERBOLA. 123 , CF-* I 1 and CE.CG=CF 2 ; because X*-Y'=l (Ex. 2). Ex. 5. If a chord of a hyperbola, be one diagonal of a parallelogram whose sides are parallel to the asymptotes, the other diagonal passes through the centre. Let the chord be PQ ; p, p the vectors to P and Q ; then t Now when one diagonal of a parallelogram is ma + n(3, the other will be ma n(3. Therefore in the case before us, the other diagonal is ! -J) And it is therefore in the same straight line with the line which joins the centre of the hyperbola with the middle point of PQ ; whence the truth of the proposition. 124 QUATERNIONS. [CHAP. VII. Ex. 6. If two tangents to a hyperbola at the extremities Qi Q' of <*> diameter, meet a tangent at P in the points T, T'; and if CD, CD' are the semi-diameters conjugate to CP, CQ ; tJten (1) PT : QT :: PT' : Q'T' :: CD : CD'- and (2) PT.PT' = CD\ If t, t', t', correspond to P, Q, Q', then ; ~^) gives t + xt = t' + x't', 1 _x 1 _ni_ ~i~~i~t'~7' t' -t x= 7^t = ~ x ' Similarly CT' = at + & + y fat - gves I y I y' __._ __ + it t' t" whence -, v r Now x : y :: x : y' gives PT : QT :: PT' : QT' :: CD : CD'. And a:?/ = 1 gives PT.PT'=CD\ COR. x'y'=l, gives QT.Q'T'=CD\ ART. 55.] THE PARABOLA AND HYPERBOLA. 125 Ex. 7. Straight lines move so that the triangular area which they cut off from two given straight lines which meet one another is constant: to find tlie locus of their ultimate intersections. Let OAA', ORE' be the fixed lines, AB, A'B'two of the moving lines with the condition that OA.OB = OA'.OB\ If a, /3 be unit vectors along OA, OB, OA = ta, OB = u p-, OA' = t' a , OK = u'p, the point of intersection of AB, AB' gives p = to. + x (u(3 to) = t'a + x' (u'(3 - t'a), .'. XU = x'u, and t (1 - x) = t' (1 - x') Now tu = t'u' = c because the triangle has a constant area; . *. x = - --, = - ultimately; t + 1 '-i the equation of a hyperbola. ADDITIONAL EXAMPLES TO CHAP. VII. 1 . In the parabola SY 2 = SP.SA. 2. If the tangent to a parabola cut the directrix in fi, SH is perpendicular to SP. 3. A circle has its centre at the vertex A of a parabola whose focus is S, and the diameter of the circle is 3AS. Prove that the common chord bisects AS. 4. The tangent at any point of a parabola meets the directrix and latus rectum in two points equally distant from the focus. 126 QUATERNIONS. [CHAP. TIL 5. The circle described on SP as diameter is touched by the tangent at the vertex. 6. Parabolas have their axes parallel and all pass through two given points. Prove that their foci lie in a conic section. 7. Two parabolas have a common directrix. Prove that their common chord bisects at right angles the line joining their foci. 8. The portion of any tangent to the parabola between tan- gents which meet in the directrix subtends a right angle at the focus. 9. If from the point of contact of a tangent to a parabola a chord be drawn, and another line be drawn parallel to the axis meeting the chord, tangent and curve ; this line will be divided by them in the same ratio as it divides the chord. 10. The middle points of focal chords describe a parabola whose latus rectum is half that of the given parabola. 11. PSQ is a focal chord of a parabola: PA, QA meet the directrix in y, z. Prove that Pz, Qy are parallel to the axis. 12. The tangent at D to the conjugate hyperbola is parallel toCP. 13. The portion of the tangent to a hyperbola which is in- tercepted by the asymptotes is bisected at the point of contact. 14. The locus of a point which divides in a given ratio lines which cut off equal areas from the space enclosed by two given straight lines is a hyperbola of which these lines are the asymp- totes. 15. The tangent to a hyperbola at P meets an asymptote in T, and TQ is drawn to the curve parallel to the other asymp- tote. PQ produced both ways meets the asymptotes in R, R : RR is trisected in P, Q. ART. 55.] THE PARABOLA AND HYPERBOLA. 127 16. From any point JK of an asymptote, UN, EM &TQ drawn parallel to conjugate diameters intersecting the hyperbola and its conjugate in P and D. Prove that CP and CD are conjugate. 17. The intercepts on any straight line between the hyper- bola and its asymptotes are equal. 18. If QQ' meet the asymptotes in R, r, 19. If the tangent at any point meet the asymptotes in X and I 7 , the area of the triangle XCY is constant. CHAPTER VIII. CENTRAL SURFACES OF THE SECOND ORDER, PARTICULARLY THE ELLIPSOID AND CONE. 56. The Ellipsoid. In discussing central surfaces of the second order, we shall speak as if our results were limited to the ellipsoid. That such limitation is not, in most cases, necessarily imposed on us, will be apparent to any one who has a slender acquaintance with ordinary Analytical Geometry. We adopt it in order that our language may have more precision, and that, in some instances, our analysis may have greater simplicity. If the centre be made the origin it is clear that the scalar equation can contain no such term as ASap, for the definition of a central sur- face requires that the equation shall be satisfied both by + p and by -p. If we turn to the equation of the ellipse (Art. 43), we shall see at once that the equation of the ellipsoid must have the form ap* + bS'ap + 2cSapSp P + ... = 1. Now if, as in the Article referred to, we put <f>p = ap + baSap + c (aS(3p + (3Sap) + ... we shall have Sp<l>p = ap* + bS*ap + 2cSapS(3p + ... -li the equation required. It will be seen that, as in Arts. 32, 33, one form of the equa- tion of the straight line was found to coincide exactly with the equation of a plane, so a form of the equation of the ellipse coincides exactly with the equation of the ellipsoid. ART. 58.] CENTRAL SURFACES OF THE SECOXD ORDER. 129 It is evident that the three properties of </a given in Art. 44 are true of </>p in its present form. 57. To find the equation of the tangent plane. Let a secant plane pass through the point whose vector is p; and let p be the vector to any point of section. Put p - p + (3, where /3 is a vector along the secant plane ; then Sp'tp = S(p Hence, observing that (44) and we have Sp'fo' = Sp<f>p + '2S(3(j>p + Now (45), as the secant plane approaches the tangent plane, the sum of these two expressions approaches in value to the first alone : that is, for the tangent plane, S{3<f>p = 0, where /? is a vector along that plane. If TT be the vector to a point in the tangent plane, .'. S (ir p) </>p = xS{3<f>p = 0, and Sir<j>p = Sp(f>p - i is the equation of the tangent plane. COR. <f>p is a vector perpendicular to the tangent plane at the extremity of the vector p. 58. If OF be perpendicular from the centre on the tangent plane; then, since <f>p is a vector perpendicular to that plane, OY- x<f>p and Sx (<f>p)* - 1, giving . or-rwrt-rl. \: Sir W. Hamilton terms <$>p the vector of proximity. [In fact vector OF T. Q. 130 QUATERNIONS. [CHAP. VIII. 59. If tangent planes all pass thnnigh a fixed point, the curve of contact is a plane curve. Let T be the fixed point ; vector a ; p the vector to a point of contact. Then (Art. 57) Sa<f>p = 1 ; i.e. Sp^a=l (44. 3), which is the equation in p of a plane perpendicular to </>a. Now </>a is the normal vector of the point where OT cuts the ellipsoid ; .. the curve of contact lies in a plane parallel to the tangent plane at the extremity of the diameter drawn to the given point. The plane of contact is called the polar plane to the point. 60. Tangent planes are all parallel to a given straight line, to find the curve of contact. Let a be a vector parallel to the given line ; then TT p + xa is a point in the tangent plane ; .'. S(p + xa) (f>p= 1 ; and Sa<f>p = 0, or /Sp(f>a = 0, the equation of a plane through the origin perpendicular to <a : that is, the curve of contact lies in a plane through the centre parallel to the tangent plane at the extremity of the diameter which is parallel to the given line. 61. To find the locus of the middle points of parallel chords. Let each of the chords be parallel to a, it the vector to the middle point of one of them j then TT + xa, tr xa are points in the ellipsoid. From the first, S(ir + xa) <f>(ir + xa) = l (Art. 56) j i. e. Sir<f>ir + 2xSrr< ART. 61.] CENTRAL SURFACES OF THE SECOND ORDER. 131 From the second, .'. subtracting, S-n-(f>a. = Q (1), i. e. the locus is a plane through the centre perpendicular to <a, or parallel to the tangent plane at the extremity A of the diameter which is drawn parallel to a. If we call this the plane BOC, B and C being any points in which it cuts the ellipsoid ; and if OB = (3, 00= y, we shall have and therefore Sa<f>(3 = 0, or a satisfies the equation >&r</3 - of the plane which bisects all chords parallel to OB (Equation 1). Let AOC be this plane which bisects all chords parallel to OB. Then, since 00 or y is a vector in it, But we have already proved that iSytfta = 0, i. e. Sa.(f>y = 0, because y is in the plane BOC ; .-. by equation (1) a, (3 both satisfy the equation of the plane Sir<j>y = 0, which is the plane bisecting all chords parallel to y ; that plane is therefore the plane AOB: we are thus presented with three lines OA, OB, OC such that all chords parallel to any one of them are bisected by the diametral plane which passes through the other two. "We may term these lines conjugate semi-diameters, and the corresponding diametral planes conjugate diametral planes. It is evident that the number of conjugate diameters is unlimited. COB. We have the following equations : (2). 92 132 QUATERNIONS. [CHAP. VIII. They shew that y is perpendicular to both <a and <f>ft, and is therefore a vector perpendicular to their plane ; hence, as in 34. 4, y = X V(}>a<j>ft. In the same way, since <y is perpendicular to both a and ft, we have or, neglecting tensors, we have the following vector equalities : y = V<j>a<}>/3, ft = F<a<y, a = V<j>ft<f>y, <y = Fa/3,. ^ft= Fay, <a = Vfty (3). Note also upon which Hamilton founded his solution of linear equations. 62. If as in Art. 47 we write \}/\(/p for <f>p, ij/p being still a vector, the equation of the ellipsoid assumes the form i. e. (44) Sif/pif/p - I (^)-=-r(^)'=-i ............ (i), which, if we put <r = ij/p, becomes To- ~ 1, the equation of a sphere. Hence the ellipsoid can be changed into the sphere and vice versd, by a linear deformation of each vector, the operator being the function i^ or its inverse. The equations now become ScuJ/ 2 ft = 0, i.e. Siffauj/ft = 0, &c., &c .................... (2). (1) and (2) shew that \[/a, ij/ft, \j/y are unit vectors at right angles to one another. If we term the sphere To- = 1 the unit-sphere, we may enunciate this result by saying that the vectors of the unit-sphere which correspond to semi- conjugate diameters form a rectangular system. ART. 63.] CENTRAL SURFACES OF THE SECOND ORDER. 133 63. Let us now take i, j, k unit vectors along the principal axes of x, y, z ; then we shall have (1), . '. Sip = x, &c. so that for the sake of transformations in which it is desirable that the form of p should be retained, we may write p = -(iSip+jSjp + kSkp) .................. (2); and as <{>p is a linear and vector function of p, its vector portions along the principal axes will be multiples of iSip, jSjp, kSkp ; we may therefore write the form a 2 having been assumed in order to make the equation Sp<f>p = 1 coincide with the Cartesian equation x 3 if z 3 __ i_ y __ L _ i a 3 + b* <r (4), we require to take \j/p so that performing the operation if/ twice on p shall give the same result (with a - sign) as performing the operation < once. Now a comparison of equations (2) and (3) will shew that the latter operation introduces -5 &c. into p ; it is evident therefore that the former operation (^) is to introduce - &c. or \ a 134 QUATERNIONS. [CHAP. VIII. It may perhaps be worth while to verify this result. We have fJStyp jSjtp kSfyp\ wilrp = I -- 1 -- - -- 1 -- \ a b c J a\ a b c / .i'Sip = t -/+... a fiSip jSjp kSkp\ ~~ + ~~^ ~ /iSip jSjp JcSkp\ "" " 1 ~' (7), because <f><j>~ l p produces p. \j/~*p = - (aiSip + bjSjp + ckSkp) ................... (8 ), (9). It is evident that the properties of Art. 44 apply to all these functions. 64. EXAMPLES. Ex. 1. Find the point on an ellipsoid, the tangent plane at which cuts off equal portions from the axes. Let x, y, z be the co-ordinates of the point, p the portion cut off, then p = xi + yj + zk. Now pi, pj, pk are points on the tangent plane ; .'. Spi<f>p= I, which gives ART. 64.] CENTRAL SURFACES OF THE SECOND ORDER. 135 or --= = 1. a Similarly -^ = 1, ? = ! x y z 1 1 a 2 b 2 c*~ p~ Jcf^t-lf + tf ' Ex. 2. To find the perpendicular from the centre of the ellipsoid on a tangent plane. 1\ 8 OY a =(T-=-\ ; (Art. 58) \ 9P/ - ^+ f! + ^ (Art. 63, 1. 3). Ex. 3. To ^/mcJ the locus of the points of contact of tangent planes which make a given angle with tlie axis ofz. "We have Z* ,/X 1 I/* Z* Or -*=P - + Ii + c \a* b c the equation of a cone whose axis is that of z and guiding curve an ellipse whose semi-axes are a 2 , b 3 . The intersection of this surface with the ellipsoid is the locus required. Ex. 4. To find the locus of a point when the perpendicular from the centre on its polar plane is of constant length. Let TT be the vector to the point, then = 1 is the equation of the polar plane (Art. 59), and T - is the length of the perpendicular on it (Art. 58) ; <f>7T 13G QUATERNIONS. [CHAP. VIII. .'. S(<j>irf = -C*, by the question. But since (44) if 8 be (f>ir, . : /Sir^ir C a is the equation required ; hence the Cartesian equation is (63. 6) T 2 II 2 Z? x >y , z r* ~i + Ti "* -- i~ ^ a b c Ex. 5. The sum of the squares of three conjugate semi-dia- meters is constant. Let a, /3, y be the semi-diameters ; i/^a, i^/3, I/Q/ are rectangular unit vectors (Art. 62). Now a = - (aiStya + bjSfya + ckSkfya) (63. 9) ; . '. (Ta) 2 = - a 2 = a 2 (Stya)" + I 2 (Sj^a) 2 + c 2 a* (StyP) 2 + VfiJtPY + c 2 a 2 (Sty?)' + b 2 (SJty) 2 + c 2 (Styy? : adding, and observing that yxtfF+(afyfF+(8fytf*i (si. cor.), we get (To) 2 + (Tpy + (Ty) 2 = a? + b 2 + c 2 , Ex. 6. The sum of the squares of the three perpendiculars from the centre on three tangent planes at right angles to one another is constant. We have p - ^- l <f>p = a'iSfyp + b 3 jSj<f>p + c 2 JcSk(j>p (63. 7), and <t>p = - (iSi^p +jSj<f>p + kSk<f> P ) (63. 2) ; .*. Sp<j>p = 1 - a 3 (SUfrp)' + b (Sjfo)' + c 2 (Sk<t>p)' {a 2 (SiUfr)* + b 2 (Sj Ufa)* + c 2 (Sk Ufa)*} ; ART. 64.] CENTRAL SURFACES OF THE SECOND ORDER. 137 hence if p, p, p" be three vectors so that <p, (f>p, <pp" are at right angles to each other ; that is, so that the tangent planes at their extremities are at right angles to one another (57. Cor.), 1 1 1 w - a 2 {(Sil7<j> P )' + (SiU^p') 2 + ( = a ' + b* + c' (31. Cor.). But , &c. are the perpendiculars from the centre on the tangent planes at p, p', p" (58). Hence the proposition. Ex. 7. The siim of the squares of the projections of three con- jugate diameters on any of the principal axes is equal to the square of that axis. Let a, ft, y be conjugate semi-diameters; then, since a = - (aiSiij/a + bjSjta + ckSktya) (63. 9), Sia. = aStya. Similarly, Si(3 = aSiij/fi, Siy aSiij/y j .-. (Sia) 3 + (Si/3) 3 + (Si 7 ) 2 = a 2 {(Stya)' + (StyP)* + (Sty?)*} = a 2 (31. Cor.), because ij/a, \f/(3, \f/y are at right angles to one another (62). But Sia is the projection of Ta along the axis of x; and similarly of the others. Hence the proposition. Ex. 8. The sum of tlie reciprocals of the squares of the three perpendiculars from the centre on tangent planes at the extremities of conjugate diameters is constant. Let Oy lt Oy a , Oy a be the perpendiculars. J-, = -(<K>* (58) (SiaY (SjaY (Ska)' ~ "~ otf.j 138 QUATERNIONS. [CHAP. VIII. i _ Oy*~ a 4 6 4 - . 4 4 4 _- 0y a " a = OSW) 2 + (Si/3? + (%)* + &c. = -,+ i+-, (Ex.7). a o~ c Ex. 9. If through a fixed point within an ellipsoid three chords be drawn mutually at right angles, the sum of tJie recipro- cals of the products of their segments will be constant. Let 6 be the vector to the given point ; a, (3, y unit vectors parallel to three chords at right angles to each other. Then 6 + xa = p gives a quadratic equation in x, the product of whose roots is -I . '. the product of the reciprocals of the segments of the chord is 1 $a<f>a 1 -1 ' (To.) 2 ' and the sum of the reciprocals of the products of the segments is (Sia)' (SjaY (Ska) 2 .. Now since SaAa = * ^ + ~r~ + r~ ( 63 - 2 J 3 )> 2 r~ 6 c the sum of the reciprocals of the products ART. 64.] CENTRAL SURFACES OF THE SECOND ORDER. 139 - ' - I \ct - Cor ')- COB. If be not constant, but S6<j>0 be so, i. e. if the given point be situated on an ellipsoid concentric with and similar to the given ellipsoid, the same is true. Ex. 10. If the poles lie in a plane parallel to yz, the polar planes cut the axis of x always in the same point. Let pi be the distance from the origin of the plane in which the poles lie, 8 any line in that plane, then ir=pi + 8 is the vector to a pole, and S P <t>(pi + o) = l (59) the equation of the corresponding polar plane. At the point where this plane cuts the axis of a?, p = xi; . . Spxi<f>i + xSi<f>8 = 1 . Now 8 is a vector in a plane perpendicular to <j>i, and Si<f>i= constant = n suppose ; .'. npx= 1, which shews that x is constant. Ex. 11. A, B and C are three similar and similarly situated ellipsoids; A and B are concentric, and C has its centre on the surface of B. To shew that the tangent plane to B at this point is parallel to the plane of intersection of A and C. Let a be the vector to the centre of C. = a the equation of A, S(p-a)<p(p-a)=c ...... (7. 140 QUATERNIONS. [CHAP. VIII. Now at the intersection of A and C, p is the same for both ; therefore the equation of the plane of intersection is to be found by subtracting the one from the other. It is therefore 2/Sp<f>a = Sat/to. + a-c ; and the equation of the tangent plane to B at the centre of C is Srr^a b ; .: both planes are perpendicular to <a, and are consequently parallel. Ex. 12. If through a given point chords be drawn to an ellipsoid, the intersections of pairs of tangent planes at their ex- tremities all lie in a plane parallel to the tangent plane at the extremity of the diameter which passes through the point. Let a be the vector to the point ; a + xfl, a + xfl, the vectors to the points of intersection with the ellipsoid of chords parallel to ft ; then STr<f> (a 4 a;^) = 1, are the equations of the tangent planes at these points. At the intersection of these planes w is the same for both ; .'. subtracting we get Sir<f>{3 = 0, STT^CL = 1 . The last equation is that of the line of intersection of the tan- gent planes; and that line is perpendicular to tf>a, or (57. Cor.) parallel to the tangent plane at the extremity of the diameter which passes through the given point. COR. S-7r<j>(3 shews that the line of intersection correspond- ing to any one chord is parallel to the tangent plane at the extremity of the diameter which is parallel to that chord. Ex. 13. Two similar and similarly situated ellipsoids are cut by a series of ellipsoids similar and similarly situated to the two ART. 64.] CENTRAL SURFACES OF THE SECOND ORDER. 141 given ones ; and in such a manner that the planes of intersection are at right angles to one another. Skew that the centres of the cutting ellipsoids lie on another ellipsoid. Let Sp<t> P = l ............................ (1), S(p-a)4>(p-a) = C ................... (2), be the given ellipsoids; S(p-ir)<}>(p-Tr} = x ...................... (3), one of the cutting ellipsoids. </> is the same for all because the ellipsoids are similar. The plane of intersection of (1) and (3) is found by subtracting the equations ; and is therefore X. The plane of intersection of (2) and (3) is + C X. The former of these planes is perpendicular to <j>ir and the latter to <f>tr <}ia ; and, since by the question, the former is perpen- dicular to the latter, <f>ir is perpendicular to </>TT </>ct, .'. S(j>ir (</>TT <a) = 0, the equation of the locus of the centres of the cutting ellipsoids. This equation will be reduced to the requisite form by ob- servin that .'. S (IT - a) < 2 7T = 0, the equation of an ellipsoid of which the semi-axes are propor- tional to a 2 , b\ c* (63. 6). The Cartesian equation is 142 QUATERNIONS. [CHAP. VIII. Ex. 14. If a tangent plane be drawn to the inner of two similar concentric and similarly situated ellipsoids the point of contact is the centre of the elliptic section of the outer ellipsoid. Let Sp<f>p - 1 be the equation of the inner, a*Sp<f>p = 1 of the outer ellipsoid. The tangent plane is STT^P = 1. Now if a- be the vector to the elliptic section measured from the point of contact, IT p + cr is a point in the outer ellipsoid ; .'. a 2 S (p + cr) < (p + cr) = 1. But crc/>p = (57. Cor.); the equation of an ellipse of which the centre is the point of contact, Ex. 15. find the equation of the curve described by a given point in a line of given length whose extremities move in fixed straight lines. First, let the straight lines lie in one plane. Let unit vectors parallel to them be a, ft. Let the vectors of the extremities of the moving line be xa, y/3, and its length I. Then the condition is or x* + y 2 + 2xySoip = l 3 (1). The vector to a point which divides this line in the ratio e : 1 is p = xa + e (yfi xa) = xa (1 - e) + ey/3 ; . '. Sap = - (1 - e) as + eySa(3, = (1 - e) xSa/3 - ey ; ART. 64.] CENTRAL SURFACES OF THE SECOND ORDER. 143 , Sap + SaBSBp SBp + SaBSao whence X ~T\ \ /tr* o i \ > y = :-, which values being substituted in equation (1) give the required equation, viz. : (Sap + SapS Pp) a + 2 a " (Sap + SapSpp) (SP P + e(L-e) = P (S 2 ap - 1)*. But p is subject to the additional condition (31. 2. Cor. 2) S . app = ; and the locus is a plane ellipse. When the given straight lines are at right angles to one another, the equation is much simplified, for O 1 O A . and our equations are x 2 + y 2 = I 2 , Sap = -(l-e)x, Sp P = -ey; whence an ellipse of which the semi-axes are le and I (1 e). Generally, if the given lines do not meet, let the origin be chosen midway along the line perpendicular to both; then we have y and y being the vectors perpendicular to the lines, P = (y+xa)(l-e) + e(-y + yp). The first gives ^ + ( Xa -y^ = -P- and the second gives, as in the simpler case above, Sap = (le)x+ eySafi, = (l-e) xSap - ey. 144 QUATERNIONS. [CHAP. VIII. Hence the elimination of x and y again leads to the equation of an ellipsoid, the only difference being that I 2 is diminished by the square of the shortest distance between the lines; i.e. the axes are less than in the former case. In the extreme case, where I = 2Ty, the equation cannot be satisfied except by x = 0, y = 0, (i. e. the locus is reduced to a single point), unless indeed we have o = *& for then x = y, and the locus is a straight line parallel to each of the preceding lines. 65. The cone. 1. To find the equation of a cone of revolution whose vertex is the origin 0. Let a be a unit vector along the axis OA, p the vector to a point P on the surface of the cone ; then Sap = Tp cos 0, being the angle POA. But this angle is constant, .*. S 2 ap - c 2 p 2 is the equation required. 2. The equation of a cone which has circular sections, but which is not necessarily a cone of revolution, is thus found. Take the vertex as the origin, and let one of the circular sections be the intersection of the plane Sap = -a' (1) with the sphere p' = Sfip (2). Since these are scalar equations we may multiply them together ; and thus obtain at all the points of the circular section 0..... (3). ART. 66.] CENTRAL SURFACES OF THE SECOND ORDER. 145 Now if xp or p be written in place of p, the equation is not changed, since p occurs twice on each side. It is therefore the required equation, of the cone. COR. 1. Every section by a plane parallel to Sap = - a 2 is a circle. For the equation of a plane parallel to Sap = a 2 is Sap aa 2 , which being substituted in the equation of the cone gives the equation of a circle. COR. 2. The plane S(3p = -bp* ........................... (4) also gives a circle whose equation is a 2 p 2 = b/3 2 Sap .......................... (5). These two equations give the subcontrary sections. To deduce the relation between the two sections ; let be the vertex of the cone, OAB the plane through a, ft; AB the line in which the section cuts this plane, AD that in which the sub- contrary section cuts it ; OA = P , OB = p, OD = xp. 6/8 2 We have, by (5), xp' 2 = ~- 8 - Sap' = P 2 ,by(2); i.e. OB:OD = OA 2 , and the triangles OAB, OAD are similar, or AD cuts OA at the same angle that AB cuts OB. 66. If <fr> - 2a 2 p + aSfip + fiSap, the equation of the cone is reduced to Sp<j>p = 0. T. Q. 10 146 QUATERNIONS. , [CHAP. VIII. It is evident that all the properties of <f>p, Ai-t. 44, are appli- cable here. As in Art. 57, the equation of the tangent plane is 0. 67. EXAMPLES. Ex. 1. Tangent planes are drawn to an ellipsoid from a given external point, to find the cone which has its vertex at the origin [the centre of the ellipsoid], and which passes through all the points of contact of the tangent planes with tlie ellipsoid. Let a be the vector to the external point, p a point in the ellipsoid where a tangent plane through a touches it. Then the equation of the ellipsoid is and the equation of the tangent plane The equation Sp<f>p = O 2 2 y / V /vO /y* ni* iy / Vy/yi If)! && \ * mf V if i <* ' ' if (/ *w<w \ rt** i " i I ^-L*^* 7 J Wl 212 2 L2 ^J a o c \a o c / represents a surface passing through the points of contact; and is the cone required. [For it is homogeneous in Tp."] Ex. 2. Of a system of three rectangular vectors two are con- fined to given planes, to find the surface traced out by the third. Let TT, p, a- be the three vectors, of which two are confined to given planes whose equations are to find the locus of <r. Since the vectors are at right angles, we have Sirp = 0, Sir(r 0, Sap = 0, and we have five equations from which to eliminate TT and p. Since SO.TT = 0, SO-TT = 0, IT is at right angles to both a and <r, and therefore to the plane off' } or IT x Fa<r. ART. 67.] CENTRAL SURFACES OF THE SECOND ORDER. 147 Since Sfip = 0, Sa-p = 0, p is at right angles to the plane /?cr; therefore and irp = xy Va<r V(3(r. Now S-rrp = 0, therefore S . Vaa- F/?<r = 0, or S (aa- - Sao-) (p<r - Spa) = 0, or o*Sap- ScurSp<r=0, the equation of a cone of the second order, which has circular sections (65. 2). COR. The circular sections are parallel to the two planes to which the two vectors are confined. Ex. 3. The equation p = t s a + u*@ + (t + uf y is that of a cone of the second order touched by each of the three planes through OAB, OBC, OCA; and the section ABC through the extremities of a, (3, y is an ellipse touched at their middle points by AB, BC, CA. 1. If the surface be referred to oblique co-ordinates parallel to a, j8, y respectively, we shall have p = xa + yfi.+ zy, therefore x = t s , y = u 2 , z = (t + u) 3 , or z = ( t jx + l jy) 2 = x + y + 2*jxy, which gives (z-x y}* = 4xy, a cone of the second order. 2. If t - u, the equation becomes p = t*(a + P), the equation of a straight line bisecting the base A, which since it satisfies the equation relative to t, shews that this line coincides with the cone in all its length; i.e. the cone is touched in this line by the plane OAB. Similarly, by putting t 0, u - respectively, we can shew that the cone is touched by the plane BOC, COA in the lines which bisect AC, CA. 102 QUATERNIONS. [CHAP. VIII. 3. Restricting ourselves to the plane ABC, we have the section of a cone of the second order enclosed by the triangle ABC, which triangle is itself the section of three planes each of which touches the cone. Ex. 4. The equation p = aa + b/3 + c-y with the condition ab + be + ca = is a cone of the second order, and the lines OA, OB, 00 coincide throughout their length with the surface. 1. It is evident that the equation gives xy + yz + zx = 0. 2. That if b = 0, c - 0, the question is satisfied by p = aa, whatever be a, therefore &c. Ex. 5. Find the locus of a point, the sum of the squares of whose distances from a number of given planes is constant. Let AS'8 ] p 1 = C' 1 , S8 2 p 2 = C 2 , &c. be the equations of the given planes, p the vector to the point under consideration; then 01,8,, x, 8 , &c. will be the perpendiculars on the planes from the point ; provided therefore SS 1 (p + oj.S,) = C l , &c. and ajjSj 2 = C l SS^, &c., <V = (C' 1 ->$V) 2 ; i.e. the square of the line perpendicular to the first plane from the given point /C.-^pV " V Z'8, ) ' and, by the question, C.-S8 lP \' /C-SS 2 p\ 2 ' - + ~ - +&c. is constant. The locus is therefore a surface of the second order. ART. 68.] CENTRAL SURFACES OF THE SECOND ORDER. 149 Ex. 6. The lines which divide proportionally the pairs of opposite sides of a gauche quadrilateral, are the generating lines of a hyperbolic paraboloid. Let ABCD be the quadrilateral. AD, EG are divided proportionally in P and R. Let CA = a CB = P, CD = ; i. e. CP y = m (a y) ; therefore RP = CP CR = = m(3 +p {y + m (a - y) ra/3} therefore x = pm, y = m pm, z = p (1 m); therefore m=x+ y> ^ = ^v x + y or (x + z) (x + y) = x, the equation referred to oblique co-ordinates parallel to a, /?, y. PASCAL'S HEXAGRAM. 68. Let be the origin, OA, OB, OC, OD, OE five given vectors lying on the surface of a cone, and terminated in a plane section of the cone ABCDEF, not passing through ; OX any vector lying on the same surface. Let OA = a, OB = fi, OC = y, OD=8, OJ = e, OX '= p. The equation S. F(FaF8e) F(F^yFep) F(FySFpa) = (1) is the equation of a cone of the second order whose vertex is and vector p along the surface. For 150 QUATERNIOXS. [CHAP. VIII. 1. It is a cone whose vertex is because it is not altered by writing xp for p. Also it is of the second order in p, since p occurs in it twice and twice only. 2. All the vectors OA, OS, OC, OD, OE lie on its surface. This we shall prove by shewing that if p coincide with any one of them the equation (1) is satisfied. If p coincide with a, the last term of the left-hand side of the equation, viz. Vpa, becomes Vaa = Va 3 = 0, and the equation is satisfied. If p coincide with ft, the left-hand side of the equation be- comes S. F(FaFSe) F(F/? r Fe) F(FySF/3a) (2). Now F(Fy3yFej3) = - F(Fe/3F/3y), (22. 2), is a vector parallel to /? (31. 3), call it mp; and F.{F(Fa/3FSe) F(FySF/2a)} = F. {F(Fa/3FSc) F(Fa/?FyS)}, (22. 2), = a multiple of Ya{3, (31. 3), = nVa{3, say. ART. 68.] CENTRAL SURFACES OF THE SECOND ORDER. 151 Hence the product of the first and third vectors in expression (2) becomes scalar + n Fa/3, and the second is m/2; therefore expression (2) becomes, by 31. 2, $ . (scalar + n Fa/3) m/2 = 0, because Fa/2 is a vector perpendicular to (3. Equation (1) is therefore satisfied when p coincides with ft. If p coincide with y both the second and third vectors are parallel to /3 (31. 3); therefore their product is a scalar, and equa- tion (1) is satisfied. The other cases are but repetitions of these. Hence equation (1) is satisfied if p coincide with any one of the five vectors a, /3, y, 8, e; i.e. OA, OB, OC, OD, OE are vectors on the surface of the cone. 3. Let F be the point in which OX cuts the plane ABCDE; then ABCDEF are the angular points of a hexagon inscribed in a conic section. 4. Let the planes OAB, ODE intersect in OP; OBC, OEF in OQ; OCD, OF A in OR', then V. Va(BV8e = mOP, (31. 4), F. V. therefore S.V(Va/3 FSe) F ( F/2y Fcp) F ( FyS Fpa) = mnpS(OP .OQ.OR}; hence equation (1) gives 8(OP.OQ.OR)=Q, or (31. 2. Cor. 2) OP, OQ, OR are in the same plane. Hence PQR, the intersection of this plane with the plane A BGDEF is a straight line. But P is the point of intersection of AB, ED, &c. 152 QUATERNIONS. [CHAP. VIII. Therefore, the opposite sides (1st and 4th, 2nd and 5th, 3rd and 6th) of a hexagon inscribed in a conic section being produced meet in the same straight line. COR. It is evident that the demonstration applies to any six points in the conic, whether the lines which join them form a hexagon or not. ADDITIONAL EXAMPLES TO CHAP. VIII. 1. Find the locus of a point, the ratio of whose distances from two given straight lines is constant. 2. Find the locus of a point the square of whose distance from a given line is proportional to its distance from a given plane. 3. Prove that the locus of the foot of the perpendicular from the centre on the tangent plane of an ellipsoid is (axy + (byy + (czy=(x* + y* + z 2 y. 4. The sum of the squares of the reciprocals of any three radii at right angles to one another is constant. 5. If Qy v Oy a , Oy a be perpendiculars from the centre on tangent planes at the extremities of conjugate diameters, and if Qu Q& Q 3 be the points where they meet the ellipsoid; then 1 1 11)1 1 . -f . = 1 1 . OY 2 00 2 OY 2 00 OY a 00 ct* b* c 6. If tangent planes to an ellipsoid be drawn from points in a plane parallel to that of xy, the curves which contain all the points of contact will lie in planes which all cut the axis of z in the same point. 7. Two similar and similarly situated ellipsoids intersect in a plane curve whose plane is conjugate to the line which joins the centres of the ellipsoids. 8. If points be taken in conjugate semi-diameters produced, at distances from the centre equal to p times those semi-diameters respectively; the sum of the squares of the reciprocals of the ART. GS.] CENTRAL SURFACES OF THE SECOND ORDER. 153 perpendiculars from the centre on their polar planes is equal to p z times the sum of the squares of the perpendiculars from the centre on tangent planes at the extremities of those diameters. 9. If P be a point on the surface of an ellipsoid, PA, PB, PC any three chords at right angles to each other, the plane ABC will pass through a fixed point, which is in the normal to the ellipsoid at P; and distant from P by 2 P !_ 1_ i' a* b 2 c 2 where p is the perpendicular from the centre on the tangent plane at P. 10. Find the equation of the cone which has its vertex in a given point, and which touches and envelopes a given ellipsoid. CHAPTER IX. FORMULAE AND THEIR APPLICATION. 69- PRODUCTS of two or more vectors. 1. Two vectors. The relations which exist between the scalars and vectors of the product of two vectors have already been exhibited in Art. 22. We simply extract them : (a) Sap = Sp a . (b) Va/3 = -Vj3a. (c) ap + pa=2Sap. (d) a/? - /3a = 2 Fa/?. These we shall quote as formulae (1). 2. We may here add a single conclusion for quaternion products. Any quaternion, such as aft, may be written as the sum of a scalar and a vector. If therefore q and r be quaternions, we may write r = Sr+Vr; qr = SqSr + SqVr + Sr Vq + Vq Vr, S . qr = SqSr +S.Vq Vr, V. qr = SqVr + SrVq + V. VqVr, where S .VqVr is the scalar part, and V.VqVr the vector part of the product of the two vectors Vq, Vr. If now we transpose q and r, and apply (a) and (b) of for- mulae 1, we get S.qr = S.rq \ V. qr + V. rq=2 (SqVr + SrVq)) " ART. G9.J FORMULA AND THEIR APPLICATION. 155 3. Three vectors. By observing that S.ySafi is simply the scalar of a vector, and. is consequently zero, we may insert or omit such an expression at pleasure. By bearing this in mind the reader will readily apprehend the demonstrations which follow, even in cases where we have studied brevity. -S.yap .............................. (3). Again, S.a{3y = S.a (S/3y + Vpy) (3). The formulae marked (3) shew that a change of order amongst three vectors produces no change in the scalar of their product, provided the cyclical order remain unchanged. This conclusion might have been obtained by a different pro- cess, thus : In (2) let q - a/2, r = y, there results at once Again in (2) let q = ya, r = ft, there results S . yap = S . Pya. We have therefore, as before, S.apy=S.yap = S.pya .................... (3). 4. S.afiy^S .aVfiy = -S.aVy(3, (by 1.6), = -S.arf ............................. (4). Similarly S . a(3y = - S . fay ........................ (4), or a cyclical change of order amongst three vectors changes the sign of the scalar of their product. 156 QUATERNIONS. [CHAP. IX. 5. It has already been seen (Art. 31. 1) that S . afiy is the volume of the parallelepiped of which the three edges which terminate in the point are the lines OA, OS, OC whose vectors are a, (3, y respectively. We may express this volume in the form of a determinant, thus : Let a, /?, y be replaced by xi +yj + zk, x'i + y'j + z'k, x"i + y"j + z"k (Art. 31. 5) ; x, y, z being the rectangular co-ordinates of A, x, y, z' those of B, x", y", z" those of C, measured from as the origin ; then S . a(3y = S . (xi + yj + zK) x (x'i + y'j + z'k) x (x"i + y"j + z"k). Now if we observe first that the scalar part of this product is confined to those terms in which all the three vectors i, j, k appear ; and secondly that the sign of any term in the product will by formulae (3) and (4) be or + according as cyclical order is or is not retained, we perceive that we have the exact con- ditions which apply to a determinant : therefore S . a{3y = - | , y , z x, y , z' .(5). r* n f* x , y , z The volume of the pyramid OABC is one-sixth of the above. Note relative to the sign of the scalar. Since ijk = - 1 (19), it is clear that if OA, OB, OC assume the positions of Ox, Oy, Oz in the figure of Art. 16, S (OA . OB . OC) will have a minus sign, whilst the order of the letters A, B, C is right-handed as seen from 0. If now we take any pyramid whatever OABC, of which the vertex is 0, and assume that S (OA . OB . OC) (which, being pro- portional to the volume of the pyramid, we may designate OABC), is negative when the order of the letters A, B, C is right-handed ART. C9.] FORMULAE AND THEIR APPLICATION. 157 as seen from 0, we shall find the following general law of signs to hold good whatever be the vertex ; viz. the sign of the scalar is minus or plus according as the order in it of the angles of the base of the pyramid is right-handed or left-handed as seen from the vertex. For example, CA BO = S (CA . CB . CO] - - Sapy = - OAC, which is plus because OABC is minus, and the order of the letters A, B, as seen from C is left-handed. 6. V.a = V. =aSpy-V.aVyp,(l.b), .b), (6). 7. V.apy = r.(Sap+ = ySafi -V. yVa/3 ; therefore F. ay+F. ya;8=2ySaj8 ....................... (7). 8. 2r.a = F. apy + V. ya{3 - ( F. ay/3 + F. ya,3) -F(aj8y + ^ay)-F(ay/? + ya^), (by 6), =F. (op + Pa)y- V. (ay + ya) p = 2ySap-2pSay, (1. c); therefore F. aFySy = 7 Sap-pSay ....................... (8). 158 QUATERNIONS. [CHAP. IX. 9. We have, by (8), therefore, by addition, V.(aVpy+pVya + yYap) = Q .............. (9). 10. F. apy = F. a (Spy + Vpy) which, by (8), = aSpy - pSay + ySap ........... (10). Another proof of this important formula is found in the identity which, by (4) and (6), is the theorem itself. 11. If in (8) we write Fa/3 in place of a, we get V. VaV = ~PS.apy ........................... (11). 12. Four vectors. If in (8) we write FaS in place of a, we obtain V(Va8Vpy) = yS.a8p-pS.a8y ............ (12). 13. By (12) we have F ( Vpy FaS) - 8S . Pya - aS . py8. But F ( Vpy FaS) - - F ( FaS Vpy). Hence, by adding the above result to (12), we get SS. pya - aS . PyS + yS . a8p - pS . aSy = 0, which, by (3) and (4), if we adopt alphabetical order, may be written aS.py8-pS.a.yS + yS.apS-8S.apy=Q ...... (13), or 8S.apy = aS.py8-pS.ay8 + yS.ap8 .......... (13), ART. CO.] FORMULA AND THEIR APPLICATION. 159 or, again, if we adopt cyclical order, aS . fty8 - 8S . afty + yS. Baft - ftS . ySa, or, finally, SS. afty = aS . ftyS-ftS.y8a + yS. Saft ........ (13). This equation expresses a vector in terms of three other vectors. The following equation expresses it in terms of the vectors which result from their products two and two. 14. F(ySa/3) may be written, first as F(y . Sa/3), and secondly as F(yS. aft), and the results compared. These forms give re- spectively V (y . Sa/3) = F. y (S. Sa/3 + F . Sa/3) = yS . a/3S + F. y (SSa/3 - aSSft + ftSSa), by (3) and (10), - yS . a/2S + VySSaft - VyaSSft + VyftSBa ; F (yS . aft) = V . (SyS + FyS) (Sap + Fa/3) - VaftSyS + VySSaft+ F. FySFa - VapSy&+ VyZSaft- F. Fa/3 FyS = Fa/3SyS + FySSa/3 - BS . afty + yS . a/3S, by (12). The two expressions being equated, and the common terms deleted, there results SS.afty= VaftSy8+ VftySaS + VyaSpS ......... (14). 15. S.afty8 = S.(S.afty+V.afty)S = S.(V.afty)5 = S . (aSfty - ftSay + ySaft) 8, by (10), = SaftSyS-SaySft$ + Sa$Sfty ............... (15). 16. S ( Vaft FyS) = S . (op - Saft) (yS - .S'yS) = SaSSpy - SaySftS, by (15) ......... (16). 17. S.aftyS=S.(Vafty)8 = S.8Vafty = S.Sap y ......................... (17). 1GO QUATERNIONS. [CHAP. IX. 18. Five vectors. As we do not purpose to exhibit any applications of the relations which exist among five or more vectors, we shall confine ourselves to simply writing down the two following expressions. F.a/?ySe = V. Sya .................... (18). 70. Many of these formulae might have been proved differ- ently, and some of them more directly, by assuming for instance that a, (3, y are not in the same plane. In this case any other vector 8 may be expressed in terms of a, (3, y, by the equation S = xa + yp + zy, (31. 5); therefore S./3yS = xS . pya = xS . a/3y, (3) , S.$a/3 = zS . yap = zS . a/3y, (3) ; therefore ?>S . a/3y = xaS . ajBy + yftS . apy + zyS . afty -- aS . PyS - pS . ySa + yS . Sa/3 which is formula 13. 71. EXAMPLES. Ex. 1. To express the relation between the sides of a spherical triangle and the angles opposite to them. Retaining the notation and figure of Ex. 2, Art. 29, we shall have Fa/3 Vpy = y' sin c . a sin a, where y, a are unit vectors perpendicular respectively to the planes OAB, OBC. Therefore F . Fa/2 F/?y = sin c sin a . /? sin E. Also -pS.apy = P sin c sin </>, (31. 1), where < is the angle between OC and the plane OAB. Now these results are equal (formula 11), therefore sin < = sin a sin . ART. 71.] FORMULAE AND THEIR APPLICATION. 161 Similarly sin (ft sin b sin A ; therefore sin a sin E = sin b sin A, or sin a : sin b :: sin A : sin B. Ex. 2. ^o find the condition that the perpendiculars from the angles of a tetrahedron on the opposite faces shall intersect one another. Let OA, OB, 00 be the edges of the tetrahedron (Fig. of Art. 31), a, /?, y the corresponding vectors. Vector perpendiculars from A and B on the opposite faces are F/3y, Fya respectively (22. 8). If these perpendiculars intersect in G, the three points A, B, G will be in one plane, whence S.(ft-a) F/3yFya = (31. 2, Cor. 2), i.e. S.(/3-a)V. FySyFya-0. Now F . Fy Fya = - yS . jSya (Formula 1 1), therefore S . (ft - a) F . Fy3y Fya = - (S/?y - Say ) S . jffya. Hence Spy = Say. Now 5(7' + a4' = -S a + a 2 = a 4- + y - = a 2 + /3 2 + y 2 - 2Say = (7-) 2 + yS' - AC* + OB*. Consequently the condition that all three perpendiculars shall meet in a point is that the sum of the squares of each pair of opposite edges shall be the same. COR. Conversely, if the sum of the squares of each pair of opposite edges is the same, the perpendiculars from the angles on the opposite faces will meet in a point. Ex. 3. If P be a point in the face ABC of a tetraliedron, from which are drawn Pa, Pb, PC, respectively parallel to OA, OB, OC to meet the opposite faces OBC, OCA, OAB in a, b, c; then will Pa Pb_ jPc OA + OJJ + ~OC~ T.Q. 11 162 QUATERNIONS. [CHAP. IX. Retaining the notation of the last examples, let OP = 8, Pa = tea, Pb = y(3, PC = 2y ; then \Jdi O " OCGL \JO == O ~~* 2//5* C/C ~- = - O ^ *Y" Now because P, A, B, C are in the same plane and because 0, a, B, C are in the same plane (2); also because O, A, b, C are in the same plane S.(S-2,/?)ya = 0, i.e. yS. fiya = S. Sya, or, by formula 3, yS.a(3y = S. Sya ........................ (3); lastly, because 0, A, JB, c are in the same plane S.(S-zy)a/3 = 0, ie. zS . yaj3 = S . 8a(3, or zS.a{3y = S. Sa(3 ........................ (4). Adding (2), (3), and (4) there results . a(3y = S. 8fiy + S. Sya+S. Sufi therefore x + y + z = I : Pa P& Pc__ O4 + OB + OC~ COB. 1. If P be in the plane ABC produced below the plane OBC, Pet as a vector will have the same sign as OA has; hence in this case we shall have _Pa L Pb Pc__ OA + OB + OC~ ART. 71.] FORMULA AND THEIR APPLICATION. 163 COR. 2. If P be outside both the planes OBC, OCA ; we shall have Pa,_Pb Pc__ ~ OA OB + 00 ~ Ex. 4. Any point Q is joined to the angular points A, B,C,0 of a tetrahedron, and ilie joining lines, produced if necessary, meet the opposite faces in a, b, c, o ; to prove that Qa Qb Qc Qo_ Aa Bb Cc Oo regard being had to the signs of Aa, Bb, &c., as in the last example. Let #4=0, QB = P, QC = y, QO = S; Qa = aa, Qb = bj3, Qc = cy, Qo = d8: then since the points a, b, c, o are in the planes 00, AGO, ABO, ABC, respectively, we have, as in the last example, aS . a (py + y8+Sp) = S. pyo, &c. &c. i.e. aS.(apy+ay8+aSp)-S.pyS = Q ............. (1), bS. (fay + (3y8 + /?8a) - S . ay8=0 ............. (2), ............. (3), Q ............ (4). Now, if we write S.apy=X, S.ay8 = y, S.aop^z, S.py8 = u; and apply the formulae 3 and 4, we get ax + ay + az u = 0, bx y bz + bu = 0, cx + cy+ z cu Q, a d which give - 1- x + -y ^ u0, ct 1 d L a 112 164 QUATERNIONS. [CHAP. IX. c b c- 1 y b-\ Z c d c-1 d-l and. therefore, 7 + -= r + , + - 7 , = 0, a- 1 6-1 c-1 tt 1 abed i.e. :r 7 =- -- , a-1 o-l c 1 a-l Qa Qb Qc Qo or ~ + "zJI + 7T + 7T = 1- A a Bb Cc (Jo Ex. 5. 7/**H>0 tetrahedra ABCD, A'B'C'D' are so situated that the straight lines, A A', BB', CC', DD' all meet in a point, the lines of intersection of the planes of corresponding faces shall all lie in the same plane. Let A' A, B'B, C'C, D'D meet in 0. The equation of the plane ABC is (34. 5) Sp ( FayS + F/3y + Fya) = S . afiy, and that of A'B'C' becomes, after dividing both sides by mnp, Sp (- VaB + - Fy + - Fya^ = S . a/?y. r \p m n ' / The vector line of intersection of the two planes is (34. 9) F. (Fa/3 + F/3y + F ya ) Q Fa^-f 1 F/3y + 1 Fya) , i.e. by formula (11), omitting the common factor S . afiy, /I 1\ /I 1\_ /I IN ( ---- a + ( ---- /3 + ----- y. \n pj \p mj \m nj From this expression the vectors of the intersections of the other planes may at once be written down. ART. 71.] FORMULAE AND THEIR APPLICATION. 165 That of ABD, A'B'D' is /I 1\ /I 1\ /I 1\_ ( --- a+ --- )/3+( --- )S; \n qj \q mj \m nj that of AC D, A'C'D'is /! l \ f l l \ /I 1\* --- )a+( --- y + ( --- )8; \p qJ \q mj \m p/ and that of BCD, B'C'D' /I 1\. /I 1\ /I 1\. --- )P + ( --- y+ --- S. \p qJ \q / ' \n p/ Now to prove that any three of these lines lie in the same plane, all that is necessaiy is to prove (31. 2, Cor. 2) that the scalar of the product of their vectors equals 0. If we take the vectors of the first three, we may write them under the form b(3 + cy, da. + b'fi + cS, a"a + b'y - bS, respectively ; so that the scalar of their product is S.(aa + bfi + cy) (a a + b'fi + cS) (a" a. + b'y - 68). Now the coefficient of every different scalar in this product is separately equal to 0. That of S . a/3y for instance is, omitting the common factor b', m \m n \p q \p m \n in which every term vanishes. That again of S . /3yB is bcb' + cb'b, which is ; and so of the rest. Hence the intersections, two and two, of the first three pairs of planes lie in the same plane ; and the same may be proved in like manner of any other three : whence the truth of the pro- position. 166 QUATERNIONS. [CHAP. IX. Ex. 6. CP, CD are conjugate semi-diameters of an ellipse, as also CP', CD' ; PP 1 , DD' are joined ; to prove that the area of tlie triangle PGP equals that of the triangle DCD'. Let a, ft, a', ft' be the vectors CP, CD, CP f , CD' ; k a unit vector perpendicular to the plane of the ellipse. Since a.*=if/~ l \l/a = (aiSi\l/a + bjSj{j/a), &c., &c. (47. 5), therefore Vaa= V. (aiStya + bjSjij/a) (aiSitya! -f bjSjij/a!) = able (Siif/aSfya SfyajSiij/a) = - abkS . k V (i/^a'). (Formula 1 6.) Similarly V/3p'=-abkS. kV(^p'). Now \l/a, \l/fi are unit vectors at right angles to one another; as are also \j/a, tyf? ; therefore the angle between iffa and tya! is the same as that between i}//3 and \frfi'. Hence S . k V (^a!) = S.kV ($&$&), and Vaa.'=Vpp, i.e. area of triangle PGP' that of triangle DCD'. Ex. 7. If a parallelepiped be constructed on the semi-con- jugate diameters of an ellipsoid, the sum of the squares of the areas of the faces of the parallelepiped is equal to the sum of the squares of the faces of the rectangular parallelepiped constructed on the semi-axes. By 63. 9, a = - (aiStya + bjSfya + c fi = - (cnStyfi + bjSfyP + c therefore Fa/3 = abk (StyaSjfyp - S + acj + bci Now Si\j/aSj^p-Si^pSj(j/a = SVij7^a, Formula (16), , (Art. 17); ART. 71.] FORMULAE AND THEIR APPLICATION. 167 therefore Va.fi = (abkSk^y + acjSfyy + bciSiij/y), Vya = - (abkStyp + acjSj^/3 + bciStyP), V(3y = (abkSktya. + acjSjif/a + bciSiij/a). If now we square and add these expressions, observing that because \f/a, if/{3, ij/y are unit vectors at right angles to one another, (Stya) 1 + (Styp)' + (Styy)* = 1, we shall have ( Fa/J)* + ( Fay)* + ( Y{3y)> = - {(ab)* + (acf + (6c) 2 }, which (21. 4) is the proposition to be proved. Ex. 8. To find the locus of t/te intersections of tangent planes at the extremities of conjugate diameters of an ellipsoid. Let TT be the vector to the point of intersection of tangent planes at the extremities of a, ft, y : then S7r</>a= 1, (57), gives Sirif/'a. = 1, or S\l/Tr\l/a.~ 1, Silnnjrp = - 1, Sif/mj/y = 1. From these three equations we extricate i}/ir by means of for- mula (14), which gives iff IT = ViJ/aif/P + Fi/f/fyy + V\fryt(/a = -3, a?_ y 2 *_ 3c? + 36 2 + 3? ~ ; an ellipsoid similar to the given ellipsoid. 1C8 . QUATERNIONS. [CHAP. IX. Ex. 9. I/O, A, B, C, D, E are any six points in space, OX any given direction, OA', OB', OC', OD', OE' the projections o/OA, OB, OC, OD, OE on OX; BCDE, CDEA, DEAB, EABC, A BCD the volumes of the pyramids whose vertices are B, C,D,E,A, with a positive or negative sign in accordance with the law given in the note to 69. 5 ; then OA'. BCDE + OB'. CDEA + OC'. DEAB + OD'. EABC Let OA, OB, OC, OD, OE be a, /?, y, S, e respectively. Write for aS (y -ft) (8- ft) (e - /?) its value a ( . ySe - S . S e/ 8 + S . c/?y - . yS), and similar expressions for /3 (a y) (S - y) (e y), die., and there will result, by addition, + cSG8-)(y-a)(8-a)=O f i.e. retaining the notation adopted in the Note referred to, OA . BCDE+ OB . CDEA + OC . DEAB + OD . EABC Now let ir be a vector along OX ; then the operation by S . TT on the above expression gives the result required. In some of the examples which follow, we will endeavour to shew how a problem should not, as well as how it should, be attacked. Ex. 10. Given any three planes, and the direction of the vector perpendicular to a fourth, to find its length so that they may meet in one point. Let /Sap = a, Sj3p = b, Syp = c be the three, and let S be the vector perpendicular to the new plane. Then, if its equation be tSSp = d, ART. 71.] FORMULAE AND THEIR APPLICATION. 169 we must find the value of d that these four equations may all be satisfied by one value of p. Formula (14) gives pS . apy = VafiSyp + VfiySap + VyaSfip by the 'equations of the first three. Operate by S . 8, and use the fourth equation, and we have the required value dS . afiy = aS . fiyo + bS . yaS + cS . afiS. Ex. 11. The sum of the (vector) areas of the faces of any tetrahedron, and therefore of any polyhedron, is zero. Take one corner as origin, and let a, ft, y be the vectors of the other three. Then the vector areas of the three faces meeting in the origin are - Fa/?, - F/3y. - Fya, respectively. a a m That of the fourth may be expressed in any of the forms lF( 7 -a)(/3-a), lF(a-/?)(y-/?), ' But all of these have the common value which is obviously the sum of the three other vector- areas taken negatively. Hence the proposition, which is an elementary one in Hydrostatics. Now any polyhedron may be cut up by planes into tetrahedra, and the faces exposed by such treatment have vector-areas equal and opposite in sign. Hence the extension. Ex. 12. If the pressure lie uniform throughout a fluid mass, an immersed tetrahedron (and therefore any polyhedron) experiences no couple tending to make it rotate. This is supplementary to the last example. The pressures on the faces are fully expressed by the vector-areas above given, and 170 QUATERNIONS. [CHAP. IX. their points of application are the centres of inertia of the areas of the faces. The co-ordinates of these points are <+0. J(/* + y)> |(y + ), l( + P + y), and the sum of the couples is I V. [Yap. (a + /3) + F/3y.(/3 + y) + Fya. (y+a) + F(y/3 + fia + ay).(a + /3 + y)} = -| F(Fa/?. y + F/?y . a+ Fya . /3) = 0, by applying formula (9). Ex. 1 3. What are the conditions that the three planes Sap = a, S(3p = b, Syp = c, shall intersect in a straight line ? There are many ways of attacking such a question, so we will give a few for practice. (a) pS . afiy = VafiSyp + VfiySap + YyaS/Bp = cVafi + aVfiy + bVya by the given equations. But this gives a single definite value of p unless both sides vanish, so that the conditions are .a/3y=0, and c Fa/? + a Vfiy + b Fya - 0, which includes the preceding. (b) S (la - m/3) p = al-bm is the equation of any plane passing through the intersection of the first two given planes. Hence, if the three intersect in a straight line there must be values of I, m such that la m(3 y, la rnh = c. The first of these gives, as before, ART. 71.] FORMULA AND THEIR APPLICATION. and it also gives Vya = m Fa/3, Vpy = -l Fa ft, so that if we multiply the second by Fa/?, la Vap - mb Fa/3 = c Fa/3 becomes a F/3y b Vya = c Fa/3 ; the second condition of (a). (c) Again, suppose p to be given by the first two in the form p = pa + qP + X Fa/3, we find a = pa* + qSap, because /So. Fa/3 = 0, 6 =pSap + qp 2 ; therefore a 8 , Sap Sap, p* a, Sap b, P 2 a , a Sap, b so that the third equation gives, operating by S . y, a 2 , Sap Sap, p a, Sap a , a Sap, b . a/3y. Now a determinate value of x would mean intersection in one point only ; so, as before, C (a*P* - S*ap) = a (p 2 Say - SapSpy) - b (SapSay- a'Spy). The latter may be written S.a[c (a/3 2 - pSap) - a (y/3 2 - pSpy) - b (aSpy - ySap)] - 0. S. a(ap 3 -pSap) = Sa(p.pa- i = -S.a(p Yap) =-S (a/3 Fa/3). Similarly, S . a (y 2 - pSpy) --= S (aft Vpy), and S . a (aSpy - ySap) = S.a(V.p Fya), (formula 8), = S (a/3 Fya). The equation now becomes S . ap (c Fa/3 + a F/3y + b Fya) = 0. 172 QUATERNIONS. [CHAP. IX. Now since S . a/3y = 0, a, (3, y are vectors in the same plane; therefore y may be written ma + nfi, and c Fa/3 4- a F/3y + b Fya assumes the form Fa/2, which, unless e = 0, gives (a/3 Fa/3) = 0, or Fa/3 is in the same plane with a, ft; but it is also perpendicular to the plane, which is absurd ; therefore e = 0, or cVa/3 + aF/3y + 6 Fya = ; thus the third and prolix method leads to the same conclusion as the first. Ex. 14. Find the surface traced out by a straight line which remains always perpendicular to a given line while intersecting each of two fixed lines. Let the equations of the fixed lines be nr = a 4- rr/3, w l = a t 4- xfi^ Then if p be the vector of the new line in any position, p = iff + y (ra'j OT) This is not, as yet, the equation required. Fur it involves essentially three independent constants, x, x lt y ; and may there- fore in general be made to represent any point whatever of infinite space. The reader may easily see this if he reflects that two lines which are not parallel must appear, from every point of space, to intersect one another. We have still to introduce the condition that the new line is perpendicular to a fixed vector, y suppose, which gives S. 7 (K 1 -vr) = Q = S. 7 [(a l -a) + x i P l -xp]. This gives x l in terms of x, so that there are now but two indeterminates in the equation for p, which therefore represents a surface, which, it is not difficult to see, is one of the second order. ART. 71.] FORMULAE AND THEIR APPLICATION. 173 Ex. 15. Find the condition that the equation s.pfr^i may represent a surface of revolution. The expression <frp here stands for something more general than that employed in Chap. VIII. above, in fact it may be written where a, a p (3, /?,, y, y t are any six vectors whatever. This will be more carefully examined in the next chapter. If the surface be one of revolution then, since it is central and of the second degree, it is obvious that any sphere whose centre is at the oi-igin will cut it in two equal circles in planes perpendicular to the axis, and that these will be equidistant from the origin. Hence, if r be the radius of one of these circles, e the vector to its centre, p the vector to any point in its circumference, it is evident that we have the following equation, where C and e are constants. This, being an identity, gives The form of these equations shews that C is an absolute con- stant, while r and e are related to one another by the first ; and the second gives (j>p Cp + e/Sep. This shews simply that . ep</>p = 0, i. e. c, p, and <p are coplanar, i. e. all the normals pass through a given straight line ; or that the expression Vp<f>p, whatever be p, expresses always a vector parallel to a particular plane. Ex. 16. If three mutually perpendicular vectors be drawn from a point to a plane, the sum of the reciprocals of the squares of their lengths is independent of their directions. 174 QUATERNIONS. [CHAP. IX. Let Sep = 1 be the equation of the plane, and let a, (3, y be any set of mutually perpendicular unit-vectors. Then, if xa, yf$, zy be points in the plane, we have = 1, ySpe = 1, zSyf = 1, whence - - oSae + pSQe + ySye (63. 2) = - + + Z . Taking the tensor, we have .i. + i + i ** * 2 2 2 x y z Ex. 17. Find the equation of the straight line which meets, at right angles, two given straight lines. Let CT = a + xft, ro- = Oj 4- a; 1 ^ 1 , be the two lines ; then the equation of the required line must be of the form and nothing is undetermined but a 2 . Since the first and third equations denote lines having one point in common, we have Similarly S . ft FySft (a, - a 2 ) = 0. Let *, = yP+yA (it is obviously superfluous to add a term in F/?ft), then s. 8.+ and, finally, Ex. 1 8. IfTp=Ta=Tp = l, and S.a/3 P = 0, ^.^(p-a)^(p- ) 8) =:/v /i(l-^). Interpret this theorem geometrically. ART. 71.] FORMULAE AND THEIR APPLICATION. 175 We have, from the given equations, the following, which are equivalent to them, p = xa Hence -x 2 -7/* + 2xySap = - 1, U( \- ( * S.U(p-a}U(p~P) -2(xy-x)Sap _x+y-\ / I -Sa/3 2 V l-x-y + xy(l+Sa(3) = x + y-l / I -Sap 2 V l-x-y + % (2xy + x* + y* - I) = x^-y-l. f~ 2 V l-2 -Sap Of course there are far simpler solutions. Thus, for instance, the given equations shew that p, a, p are radii of some unit circle. Hence the expression is the cosine of the supplement of the angle between two chords of a circle drawn from the same point in the circumference. This is obviously half the angle 176 QUATERNIONS. [dlAP. IX. subtended at the centre by radii drawn to the other ends of the chords. The cosine of this anle is and therefore the cosine of its half is v^ j Ex. 19. Find the relative position, at any instant, of two points, which are moving uniformly in straight lines. If a', ft be their vector velocities, t the time elapsed since their vectors were a, ft, their relative vector is p = a + t of - ft - tft' so that relatively to one another the motion is rectilinear, and the vector velocity is a'-/?. To find the time at which the mutual distance is least. Here we may write Tp* = -y 2 -2tSyo-t>? As the last term is positive, this is least when it vanishes, Le. when t = -S.y8- 1 . This gives p = y S/S'yS" 1 = 7 F8-'y, the vector perpendicular drawn to the relative path; as is, of course, self-evident. Ex. 20. Find the locus of a given point in a line of given length, when the extremities of the line move in circles in one plane. (Watt's Parallel Motion.) ART. 71.] FORMULA AND THEIR APPLICATION. 177 Let a- and r be the vectors of the ends of the line, drawn from the centres a, ft of the circles. Then if p be the vector of the required point subject to the conditions From these equations o- and T must be eliminated. We leave the work to the reader. There is obviously an equation of con- dition S.y(ft-a) = Q. Ex. 21. Classify the curves represented by an equation of the form a + xft + x s y P a + bx + cx* ' where a, ft, y are given vectors, and a, b, c given scalars, In the first place we remark that x 2 in the numerator merely adds a constant vector to the value of p, unless c = 0. Thus, if c do not vanish, the equation may be written, with a change of a and ft and in general a change of origin, a + xft a + bx + ex* ' and this again, by change of x and of a and ft, as a + xft It is obvious that this represents a plane curve. . , Sap a 2 + xSaft ^3p = Xaft+xft*' T. Q. 12 178 QUATERNIONS. [CHAP. IX. Hence both numerator and denominator of x are of the first degree in Sap, S(3p ; and therefore bap = car gives an equation of the third degree in p by the elimination of x. When we have Sa(3 = 0, a 2 Sap = a + ex whence - , p /oap and a (Sap)* + c ^ (Spp}* = a* Sap, a conic section. If c = 0, then with a change of x, a, ft, y, the equation may be written a hyperbola so long at least as 6 does not also vanish. If 6 and c both vanish, the equation is obviously that of a parabola. If a and 6 both vanish, whilst c has a real value, we have again a parabola. If a vanish while 6 and c have real values, we have again a hyperbola. Ex. 22. Find the locus of a point at which a given finite straight line subtends a given angle. ART. 71.] FORMULAE AND THEIR APPLICATION. 179 Take the middle point of the line as origin, and let a be the vectors of its ends. At p it subtends an. angle whose cosine is This, equated to a constant, gives the locus required. We may write the equation This is, obviously, a surface of the fourth order; a ring or tore formed by the rotation of a circle about a chord. When c = 0, i. e. when the angle is a right angle, the two sheets of this surface close up into the sphere A plane section (in the plane a, ft (suppose) where T(3 = and Sa(3 = 0) gives p = xa + yfi, (a* (1 - X s ) - yV} 2 = c 2 {(x - I) 2 + y 2 } {( + I) 2 + y 3 } a*, o r {1 - (x 2 + y 2 )} 2 - c 2 {( + y* + I) 2 - 4s 2 }, or, finally, 1 - (x* + /) = * -= , which, of course, denotes two equal circles intersecting at the ends of the fixed line. Ex. 23. A ray of light falls on a thin reflecting cylinder, shew that it is spread over a right cone. Let a be the ray, T a normal to the cylinder, p a reflected ray, /3 the axis of the cylinder. Then T is perpendicular to /?, or S(3r = Q .............................. (1). Again p and a make equal angles with T, on opposite sides of it, in one plane ; therefore p||TttT or V '. TOT/) = ........................... (2). 122 180 QUATERNIONS. [CHAP. IX. Eliminating T between (1) and (2) we have a 2 \Sap the equation of the right cone of which (B is the axis, and a a side. ADDITIONAL EXAMPLES TO CHAP. IX. 1. Prove that S . (a + )8) (/3 + y) (y + a) = 2S . a/3y. 2. S . Fa/3 F/3y Fya = - (Sa/3y) 2 . 3. S . F ( Fa/3 F/3y) F ( F/3y Fya) F ( FyaFa) = -(S. a/3 7 ) 4 . 4. ^ ( F/?y Fya) = y 2 ^a/3 - SfiySya. /3 2 (Sya) 7. 8. (a^y) 2 - o'jS V + 2ayS . a^y. 9. ^ ( Fay Fj3yo Fyay8) = ISafiSpySyaS . afiy. 10. The expression Fa/3 FyS + Fay F8^ + FaS F/3y denotes a vector. What vector 1 (Tait's Quaternions. Miscellaneous Ex. 1.) 1 1 . SapS . yS - SppS . ySa + SypS . Sa/3 - SSpS . a/3y = 0. 12. (a^y) 2 = 2a 2 /3 2 y 2 + a 2 (/3y) 2 + /3 2 (ay) 2 + y 2 (a^) 2 - ay$aj3S(3y. (Hamilton, Elements, p. 346.) ART. 71.] FORMULA AND THEIR APPLICATION. 181 13. With the notation of the Note, Art. 69. 5, we shall have DABC =OABC- OBCD + OCDA - ODAB. 14. When A, B, (7, D are in the same plane, a.BCZ>-p.CDA+y.DAB-S.ABC = 0, where BCD, &c. are the areas of the triangles. 15. SF. afiy + aV.ftyS + (3V. ySa + yF. Sa/? = 4S. a/fyS. 16. Va/3 FyS + F/3y FSa + FyS FayS + VSa V(3y is a scalar. What is its geometrical meaning ? 17. Find the equation of the sphere circumscribing a given tetrahedron. 18. A straight line intersects a fixed line at right angles, and turns uniformly about it while it slides uniformly along it. Find the equation of the surface described (1) when the fixed line is straight, (2) when it is circular. CHAPTER X. VECTOR EQUATIONS OF THE FIRST DEGREE. WITH the object of giving the student an idea of one of the physical applications of Quaternions, we will treat the solution of linear and vector equations from an elementary kinematical point of view. For this purpose we choose the problem of the de- formation of a solid or fluid body, when all its parts are similarly and equally deformed. DBF. Homogeneous Strain is such that portions of a body, originally equal, similar, and similarly placed, remain after the strain equal, similar, and similarly placed. Thus straight lines remain straight lines, parallel lines remain parallel, equal parallel lines remain equal, planes remain planes, parallel planes remain parallel, and equal areas on parallel planes remain equal. Also the volumes of all portions of the body are increased or diminished in the same proportion, as is easily seen by supposing the body originally divided into small equal cubes by series of planes perpendicular to each other. After the strain, these cubes are all changed into similar, similarly placed, and equal parallelepipeds. It is thus obvious that a homogeneous strain is entirely deter- mined if we know into what vectors three given (non-coplanar) vectors are changed by it. Thus if a, ft, y become a', ft', y CHAP. X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 183 respectively: any other vector, which may of course be expressed as p= * (aS is changed to 1 ,M p = , Q- (a- *. b.aBy^ No needful generality is lost, while much simplification is gained, by taking a, B, y as unit vectors at right angles to one another. This is, in fact, the method already spoken of, i. e. the imaginary division of the body into small equal cubes, by three mutually perpendicular series of equidistant planes. We thus have p = - (aSap + BSBp + ySyp), p' = - (a' Sap + B'SBp + y'Syp), Comparing these expressions we see that Homogeneous Strain alters a vector into a definite linear and vector function of its original value. In abbreviated notation, we may write (as in Art. 63, though our symbol, as will soon be seen, is more general than that there employed) <f>p = (a Sap + B'SBp + y'Syp), where < itself depends upon nine independent constants involved in the three equations <f>a = a' I <t>y = y i For a', B', y may of course be expressed in terms of a, B, y : and, as they are quite independent of one another, the nine co- efficients in the following equations may have absolutely any values whatever ; <f>a = a Aa. + cB + b'y ] ^ <f>y = y = b a + a'B + Gy) 184 QUATERNIONS. [CHAP. In discussing the particular form of <j!> which occurs in the treatment of central surfaces of the second order we found, Art. 44, that it possessed the property S . cr(j)p = S. p<f><r ......................... (&), whatever vectors are represented by p and o-. Remembering that a, (3, y form a rectangular unit system, we find from (a) with other similar pairs ; so that our new value of < satisfies (&) if, and only if, we have in (a) c = c The physical meaning of this condition, as will be seen im- mediately, is that the distortion expressed by < takes place without rotation. In this case the nine constants are reduced to six. But, although (6) is not generally true, we have S.<r<l>p = - (Sa'aSap = -S.p where the expression in brackets is a linear and vector function of o-, depending upon the same nine scalars as those in </> ; and which we may therefore express by < 7 , so that <}>'<T = -(aSa'<T + pSp'(r + ySy'<r) ............... (d). And with this we have obviously S . a-(f>p = S . p<'<r ......................... (e), which is the general relation, of which (6) is a mere particular case. By putting a, ft, y in succession for o- in (d) and referring to (a) we have tj/y = b'a + aft + Cy> X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 185 Comparing (/) with (a) we see that $P = <f>'p> whatever be p, provided the conditions (c) be fulfilled. This agrees with the result already obtained. Either of the functions < and <', thus defined together, is called the Conjugate of the other : and when they are equal (i. e. when (c) is satisfied) < is called & Self-Conjugate function. As we employed it in Chap. VI, < was self-conjugate-; and, even had it not been so, it was involved (as we shall presently see) in such a manner that its non-conjugate part was necessarily absent. We may now write, as before, <j>p = ('a Sap + P'S/3p + y'Syp), and, by (d), <>'p = (aSa'p + PSft'p + ySy p). From these we have by subtraction, ((/> <') p = <p <'p = aSa'p a'Sap + (3S{3'p fi'Sflp + y^j'p - V . Vaa'p + V. Vp/3'p + V . Vyy'p = 2F.ep .................................................... (y); if we agree to write We may now express that < is self-conjugate by writing e = 0, the physical interpretation of which equation is of the highest importance, as will soon appear. If we form by means of (a) the value of c as in (h) we get 2c = (cy - 6'j8) + (ao - c'y) + (bj3 - a a) which obviously cannot vanish unless (as before) the three con- ditions (c) are satisfied. 186 QUATERNIONS. [CHAP. By adding the values of <p and <'p above we obtain (< + <') p=<]>p + (j>'p = - (aSa'p + a'Sap +(3S(3'p +(3'Spp +ySy'p+y'Syp) = - F (apa' + (3pp + ypy') - p (Saa + Sftft' + Syy). As we have (by 69. 6) V . apa! = F . a'pa, &C. this new function of p is self-conjugate. This will easily be seen by putting < + <' for (p in (b) and re- membering that (by 69. 17) we have S . crapa = S . pa'cra = S . pacra', &c., &c. Hence we may write (< + (') p = 2arp ........................ (i), where the bar over OT signifies that it is self-conjugate, and the factor 2 is introduced for convenience. From (gr) and (i) we have ,/ - r r | .......................... <p p 'ufp V f.p) If instead of <f>p in any of the above investigations we write (< + g) p, it is obvious that <j>p becomes (</>' + g)p: and the only change in the coefficients in (a) and (/") is the addition of g to each of the main series J, B, C. "We now come to Hamilton's grand proposition with regard to linear and vector functions. If < be such that, in general, the vectors p, <p, <f>"p (where < 2 p is an abbreviation for < (<p)) are not in one plane, then any fourth vector such as < 3 p (a contraction for < (<(<p))) can be expressed in terms of them as in 31. 5. Thus fj> 8 p = in 3 <j> 2 p m^p + mp .................. (&)> where m, m l , m 3 are scalars whose values will be found immedi- ately. That they are independent of p is obvious, for we may put X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 187 a, ft, y in succession for p and thus obtain three equations of the form tj) 3 a = m 2 <f) 2 a m i (j)a + ma (I), from which their values can be found. For by repeated applica- tions of (a) we can express (I) in the form Aa + p + Cy = 0. This gives A = 0, $=0, C = 0. These are three equations connecting m, m^ m 2 , with the nine coefficients in (a). The other two groups of three equations, furnished by the other two equations of the form (?), are merely consistent with these ; and involve no farther limitations. This method, however, is very inferior to one which will shortly be given. Conversely, if quantities m, m l , m a can be found which satisfy (I), we may reproduce (&) by putting p = xa + yf$ + Zy and adding together the three expressions (I) multiplied by x, y, z respectively. For it is obvious from the expression for <f> that X<j>p e < (xp), X(f>*p = < 2 (xp), &C., whatever scalar be represented by x. If p, <p, and < 2 p are in the same plane, then applying the strain < again we find tj>p, (f> 2 p, < 3 p in one plane ; and thus equa- tion (k) holds for this case also. And it of course holds if <j>p is parallel to p, for then <jfp and <j> 3 p are also parallel to p. We will prove that scalars can be found which satisfy the three equations (F) (equivalent to nine scalar equations, of which, however, as we have seen, six depend upon the other three) by actually determining their values. The volume of the parallelepiped whose three conterminous edges are X, p., v is (31. 1) S . X.v. 188 QUATERNIONS. After the strain its volume is so that the ratio [CHAP. S . S . ,, . iO . A/JiV is the same whatever vectors X, //,, v may be ; and depends there- fore on the constants of </ alone. We may therefore assume and by inspection of (k) we find >v S . S . which gives the physical meaning of this constant in (/<;). As we may put if we please we see by (a) that S . (ba. m = S.afy A, c, b' c', B, a b, a', C which is the expression for the ratio in which the volume of each portion has been increased. This is unchanged by putting <' for </>, for it becomes, by (/), m - ' A, c', b c f , a' b', a, C Hence conjugate strains produce equal changes of volume. Recurring to (m) we may write it by (e) as S . A X.] VECTOR EQUATIONS OF T1IE FIRST DEGREE. 189 from which, as X is absolutely any vector, we have or . =.^ , ^ V(j>fji<f>'v = m Vfj(.v) [In passing we may notice that (n) gives us the complete solution of a linear and vector equation such as <<r= 8, where 8 and <j* are given and cr is to be found. We have in fact only to take any two vectors //. and v which are perpendicular to 8, and such that F/zv = S, and we have for the unknown vector <r = m which can be calculated, as < is given.] If in (n) we put < + g for <f> we must do so for the value of m in (m). Calling the latter N g we have S.(<j>+g)\ 9* S . \fJLV S . Xp,<j)v + S . vXtfrfj. + S . fj.v(f)X /S . and by (n) (<f> + g} V (<' + g) n(<j>' + g) v = M,. V^v ......... (p), or /v 'v) + g* F/*v] - M g From the latter of these equations it is obvious that must be a linear and vector function of F/xv, since all the other terms of the equation are such functions, 190 QUATERNIONS. [CHAP. As practice in the use of these functions we will solve a problem of a little greater generality. The vectors Vpv, F</)'/AV, and V^'v are not generally coplanar. In terms of these (31. 5), let us express Let Operate by S . A, S . p, S . v successively, then S . fj.v(j)'X = xS . AJU.V + yS . vXfip. + zS . S . {j.v<f>'fji, yS '. v (*.$'[*-, S . [J.V(j>'v = zS . VfJ.(f>V. The two last equations give (by 69. 4) y = -l, = -!, and therefore the first gives $ . Jivt'X + S . v\(> ' JL 4- S . Hence, finally, <F/x.v = /i s F/iv V(f>' IJLV Vnfiv ............... (r). Substituting this in (q), and putting tr for F/X.V, which is any vector whatever, we have (< + 9) [<t>~* +ff(^- ^)+ff 2 ]^ = (m + P- 1 ff + ^ 2 ff i + Sf 3 ) <r, or, multiplying out, (m - g<f + ii a g<t> - g*<f> + gm^T l +g a <f t + g*^ + g 3 ) <r that is (- <jt s + p a <j> + m<l>~ 1 ) <r = /^o-, or (<#> 3 -fi 2 ^ a + /i 1 <)!)-m)o-=0. Comparing this with (k) we see that S . \u.d>v + S . vX<t>u. + S . u.vd>\ = = - y O . AfJLV i 1 A})' . AylAV and thus the determination is complete. X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 191 We may write (k), if we please, in the form m<p~ l p = m,p - m 2 <f>p + <p a p (&'), which gives another, and more direct, solution of the equation (above mentioned) (f)(T = 8. Physically, the result we have arrived at is the solution of the problem, " By adding together scalar multiples of any vector of a body, of the corresponding vector of the same strained homo-* geneously, and of that of the same twice over strained, to repre- sent the state of the body which would be produced by supposing the strain to be reversed or inverted." These properties of the function < are sufficient for many applications, of which we proceed to give a few. I. Homogeneous strain converts an originally spherical por- tion of a body into an ellipsoid. For if p be a radius of the sphere, tr the vector into which it is changed by the strain, we have o- = p, and Tp = C, from which we obtain jtyrvc, or tf.^-'o^-W-C", or, finally, 8 . ox^c/TV - - C\ This is the equation of a central surface of the second degree ; and, therefore, of course, from the nature of the problem, an ellipsoid. II. To find the vectors whose direction is unchanged by the strain. Here <p must be parallel to p or <t>p=gp. This gives <f> 2 p = g 2 p, &c., 192 QUATERNIONS. [CHAP. so that by (k) we have g 3 - m 2 g 2 + m$ -m=0. This must have one real root, and may have three. Suppose g l to be a root, then <t>P ~ 9iP = > and therefore, whatever be A, S\({ip gfiXp 0, or S.p(^'\-g l \) = 0. Thus it appears that the operator <' g l cuts off from any vector A. the part which is parallel to the required value of p, and there- fore that we have where is absolutely any vector whatever. This may be written as (m t> ^t/ 1 The same result may more easily be obtained thus : The expression (< 3 - mtf + mrf -m)p = 0, being true for all vectors whatever, may be written (4>-<7 1 )(</>-<7 2 )(<-<7 3 )P = 0> and it is obvious that each of these factors deprives p of the por- tion corresponding to it : i. e. < g l applied to p cuts off the part parallel to the root of (< - grj o- = 0, &c., &c. so that the operator (<J> g y ) (<f> ~ g^) when applied to a vector leaves only that part of it which is parallel to or where X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 193 III. Thus it appears that there is always one vector, and that there may be three vectors, whose direction is unchanged by the strain. DEF. Pure, or non-rotational, strain consists in altering the lengtlis of three lines at right angles to one anotJier, without altering tlieir directions. Hence if = the strain < is pure if, and not unless, p,, p 2 , p 3 form a rectangular system. [There is a qualification if two or more of ff l ff s g 3 be equal.] Hence, for a pure strain, we have and or S Pl <f>p 9 = S Pa fa. But we have, generally, As we have two other pairs of equations like these, we see that < = <' when the strain is pure. Conversely, if <f> = <j the three unchanging directions p l} p a) p s are perpendicular to one another. For, in this case, the roots of Jf,0 are real. Let them be such that Cf-*)ft~0] (<*>-?,) p,=o[, (*-fc)iy-OJ T. Q. 13 194- QUATERNIONS. [CHAP. tnen (because, by hypothesis, the strain is pure) for <t>Pz = 92P 2 and <p'p a = g a P a - Hence, except in the particular case of ffi = ff*> we must have 8piP, = > whence the proposition. When <7, and g z are equal, p l and p 2 are each perpendicular to p a , but any vector in their plane satisfies <Jb(T gTjO- 0. When all three roots are equal, every vector satisfies <fxr - gp = 0. IV. Thus we see that when the strain is unaccompanied by rotation the three values of g are real. [But we must take care to notice that the converse does not hold. This will be discussed later.] If these values be real and different, there are three vectors at right angles to one another which are the only lines in the body whose directions remain unchanged. When two are equal, every vector parallel to a given plane, and all vectors perpendicular to it, are unchanged in direction. When all three are equal no vector has its direction changed. "V. There is, however, a peculiarity to be noticed, which dis- tinguishes true physical strain from the results of our mathe- matical analysis. When one or more of the values of g has a negative sign, we cannot interpret physically the result without introducing the idea of a pure strain which shall, as it were, pull the parts of an originally spherical portion of the body through the centre of the sphere, and so form an ellipsoid by turning a part of the body outside in. When two, only, are negative we X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 195 can represent physically the result by introducing the conception of a rotation through two right angles about the third axis. But we began by assuming that there is no rotation ! Hence, for the case considered, all three roots must be positive. See end of next section (VI.). VI. This will appear more clearly if we take the case of a rigid body, for here we must have, whatever vectors be repre- sented by p and cr, Spcr = S . i. e. the lengths of vectors, and their inclinations to one another, are unaltered. In this case, therefore, the strain can be nothing but a rotation. It is easy to see that the second of these equa- tions includes the first; so that if, for variety, we take < as represented in equations (a), and write yft + zy, we have, for all values of the six scalars x, y, z, g, 77, , the follow- ing identity : '2 / O'2 '2 ^ + (xr} + y) So! ft + (y + mi) Sfty' + ( + *) Sy'a. This necessitates i.e. the vectors a', ft', y form, like a, ft, y, a rectangular unit system. And it is evident that any and every such system satisfies the given conditions. But the system a', /8', y' must be similar to a, ft, y, i e. if a quadrant of positive rotation round a changes ft to y &c. a quadrant of positive rotation about a must change ft' to y' &c. When this is not the case, the system a, ft', y is the per- 132 196 QUATEENIONS. [CHAP. version of a, /?, y, i. e. its image in a plane mirror ; and the strain is impossible from a physical point of view. This is easily seen from another point of view. The volume of the parallelepiped whose edges are rectangular unit vectors a, (3, y is S . afiy if a positive quadrant of rotation round a brings (3 to coincide with y &c. But, in the perverted system, the volume has changed sign and is expressed by 8.* fa. VII. It may be interesting to form, for this particular case, the equation giving the values of g. We have W _S.(<f> + g)a (< g " S.afiy S.afiy Recollecting that a, ft, y ; a', /?', y are systems of rectangular unit vectors, we find that this may be written Hence the roots of M g = are in this case ; first and always, ?'=-*! which refers to the axis about which the rotation takes place secondly, the roots of Now the roots of this equation are imaginary so long as the coefficient of the first power of g lies between the limits * 2. Also the values of the several quantities W, S(3/3', Syy can never exceed the limits 1. When the system a, yS, y coincides X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 197 with a', ft', y', the value of each of the scalars is 1, and the coefficient of the first power of g is + 2. When two of them are equal to + 1 and the third to - 1 we have the coefficient of the first power of g = 2. These are the only two cases in which the three values of g are all real. In the first, all three values of g are equal to 1, i. e. <1>P = P for all values of p, and there is no rotation whatever. In the second case there is a rotation through two right angles about the axis of the - 1 value of g. VIII. It is an exceedingly remarkable fact that, however a body may be homogeneously strained, there is always at least one vector whose direction remains unchanged. The proof is simply based on the fact that the strain-function depends on a cubic equa- tion (with real coefficients) which must have at least one real root. IX. As an illustration of what precedes (though one which must be approached cautiously), suppose a body to be strained- so that three vectors, a", (3", y" (not coplanar, and not necessarily at right angles to one another), preserve their direction, becoming e^", e a ft", e a y". Then we have <f> P S . a"ft"y" = e^'S. ft"y"p + e"S . y"a"p + e a y"S . a" ft" p. By the formulae (m, s) we have S (af'+PW = ~ S (a'P'ty" + ft"y"<}>a" + y'a'Aft") nravyr -= e ^ e * +e *> so that we have by (k) (^- 1 )(^- a )(4-?Jp=0. Though the values of g are here all real, we must not rashly adopt the conclusions of (iv.), for we must remember that a", j8", y" do not, like a, ft, y, necessarily form, a rectangular system. 198 QUATERNIONS. [CHAP. In this case we have #pS . a"ft"y" = e r Vfi'fSa'p + e g Vy"a"Sft"p + eja." ft" Sy" p. So that, by (7i), 2e . a."ft"y" = V. (e,a" Vp'y" + eft" Vy"a" + e.y" Va"ft") This vanishes, or the strain is pure, if either 1. So." ft" = Sft"y" m Sy"a" = 0, Le. if a", ft", y" are rectangular, in which case e^ e 2 , e a may have any values ; or 2. e l = e a = e 3 , in which case #pS. a"ft"y" = 6l { Vft"y"Sa"p+Vy"a"Sft"p + Va"ft"Sy"p} = e lP S.a"ft"y" by (69. 14), so that ftp = e t p = <f>p for every vector : a general uniform dilatation unaccompanied by change of direction. 3. e l = e 2 , and a" and ft" both perpendicular to y". From what precedes it is evident that for the complete study of a strain we must endeavour to distinguish in each case between the pure strain and the merely rotational part. If a strain be capable of being decomposed into 1st a pure strain, 2nd a rotation, it is obvious that the vectors which in the altered state of the body become the axes of the strain- ellipsoid (i.) must have been originally at right angles to one another. The equation of the strain-ellipsoid is and in this it is obvious that (f>~* is self-conjugate, or at least is to be treated as such : for a non-conjugate term in <~ 2 /a would be (y) of the form Vep, and would therefore not appear in the equation. X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 199 For the proper treatment of rotations, the following simple but excessively important proposition, due to Hamilton, forms the best starting-point. If q be any quaternion, the operator q ( ) q~ l turns the vector, quaternion, or body operated on round an axis perpendicular to the plane of q and through an angle equal to double that of q. For the proof we refer the reader to Hamilton's Lectures, 282, Elements, 179 (1), or Tait, 353. It is obvious that the tensor of q may be taken to be unity, i. e. q may be considered as a mere versor, because the value of its tensor does not affect that of the operator. [A very simple but important example of this proposition is given by supposing q and r to be both vectors, a and fi let us say. Then is the result of turning /? conically through two right angles about a, i. e. if a be the normal to a reflecting surface and (3 the incident ray, a/3a -1 is the reflected ray.] Now let the strain <j> be effected by (1), a pure strain & (self- conjugate of course) followed by the rotation q ( ) q~\ We have, for all values of p, whence <p'p = 5r (q~ l pq). The interpretation is that, under the above definition, the con- jugate to any strain consists of the reversed rotation, followed by the pure strain. We may of course put, as in Chap, vi, vHp ejuSap + e^ftSfip + e 3 ySyp, where a, ft, y form a rectangular system. Hence <pp = e^aq^Sap + 200 QUATERNIONS. [CHAP. Here the axes are parallel to qaq~\ q(3q~ l , qyq~ l , and we have S. qaq~*qpq~ l = S . qa(3q~ l = Sa{3 = 0, &c. So far the matter is nearly self-evident, but we now come to the important question of the separation of the pure strain from the rotation. By the formulae above we see that so that we have in symbols, for the determination of CT, the equation <f) <f) = W . That is, as we see at once from the statements above, any strain, followed by its conjugate, gives a pure strain, ivhich is the square (or the result of two applications) of the pure part of either. To solve this equation we employ expressions like (&). <ft'<f> being a known function, let us call it w, and form its equation as w 3 m 2 w 2 + nijto m = 0. Here the coefficients are perfectly determinate. Also suppose that the corresponding equation in OT is ^-g^+g^-g^O, where g, g^ , g 2 are unknown scalars. By the help of the given relation TO* = w, we may modify this last equation as follows : whence = - *- X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 201 i. e. tzr is given definitely in terms of the known function <o, as soon as the quantities g are found. But our given equation may now be written or w 3 - < - 2< a, 2 + - 2 o - = 0. As this is an equation between w and constants it must be equivalent to that already given : so that, comparing coefficients, we have 9* = m; from which, by elimination of g and g s , we have The solution of the problem is therefpre reduced to that of this biquadratic equation ; for, when g l is found, g a is given linearly in terms of it. It is to be observed that in the operations above we have not been particular as to the arrangement of factors. This is due to the fact that any functions of the same operator are commutative in their application. Having thus found the pure part of the strain we have at once the rotation, for (v) gives ^-'p^qpq- 1 , or, as it may more expressively be written, If instead of (v) we write 202 QUATERNIONS. [CHAP. we assume that the rotation takes place first, and is succeeded by the pure strain. This form gives and whence to is found as aboA'e. And then (v r ) gives 5Tty = r( )r-\ Thus, to recapitulate, a strain < is equivalent to the pure strain *J <!>'<$> followed by the rotational strain <f> /^ , or to the ' rotational strain -.- _ . < followed by the pure strain J <$>$'. This leads us, as an example, to find the condition that a given strain is rotational only, i.e. that a quaternion q can be found such that Here we have <' = q~ l ( ) q, or <' = <~ 1 But m^)" 1 = T/IJ - m a <l> or mfi = ?, whose conjugate is m< = m, 7 and the elimination of <' between these two equations gives + < 2 ) + a = (m t m l mm^y + m (m a mm* + 2m 1 jM^ + m*) (m 3 mm* + Zm l m i m) 4> + (2m l + m* by using the expression for < 4 from the cubic in X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 203 Now this last expression can be nothing else than the cubic in </> itself, else </> would have two different sets of constants in the form (&), which is absurd, as these constants, from, the mode in which they are determined, can have but single values. Thus we have, by comparing coefficients, m a a = 2m l + m 2 a - mm a m t mm. m 3 mm* + 2m.m., m | I * mm = m?rn mm + m* The first gives m l = mm a , by the help of which the second and third each become m 3 - m = 0. The value m = is to be rejected, as otherwise we should have been working with non-existent terms ; and m, as the ratio of the volumes of two tetrahedra, is positive, so that finally m 1, m 1 = m a , and the cubic for a rotational strain is, therefore, or > where m is left undetermined. By comparison with the result of (vn.) we see that in the notation there employed The student will perhaps here require to be reminded that in the section just referred to we employed the positive sign in operators such as < + g. In the one case the coefficients in the cubic are all positive, in the other they are alternately posi- tive and negative. The example we have given is a particularly valuable one, as it gives a glimpse of the extent to which the 204 QUATERNIONS. [CHAP. separation of symbols can be safely carried in dealing with, these questions. DEP. A simple shear is a homogeneous strain in which all planes parallel to a fixed plane are displaced in the same direction parallel to that plane, and therefore through spaces proportional to their distances from that plane. Let a be normal to the plane, /? the direction of displacement, the former being considered as an unit- vector, and the tensor of the latter being the displacement of points at unit distance from the plane. We obviously have, by the definition, Sap = 0. Now if p be the vector of any point, drawn from an origin in the fixed plane, the distance of the point from the plane is Sap. Hence, if o- be the vector of the point after the shear, This gives <j>'p = p which may be written as = P -Tp.aS. so that the conjugate of a simple shear is another simple shear equal to the former. But the direction of displacement in each shear is perpendicular to the unaltered planes in the other. The equation for <f> is easily found (by calculating m, m 1} m a from (m), (s)) to be < 3 -3< B + 3e-l=0. Putting <'< = \J/, we easily find (with b = T/3) ^ 3 _ (3 + b 2 ) ^ + (3 + b 2 ) $- 1 = 0. Solving by the process lately described, we find X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 205 If b = 2, this gives ^ = 1, and the farther equation ^ + ^'-13^-21 = 0, of which y l 3 is a root, so that *'-4r>-r-4 and g v = I 2 J2. We leave to the student the selection (by trial) of the proper root, and the formation of the complete expressions for the pure and rotational parts of the strain in this simple and yet very interesting case. As a simple example of the case in which two of the roots of the cubic are unreal, take the vector function when the strain is equivalent to a rotation about the unit vector a ; the others of the rectangular system being /?, y. Here we have, obviously, <f>a = a, </? = (3 cos + y sin 9, <j>y = y cos - (3 sin 0, whence at once - <p = aSap + (ft cos + y sin 6) S(3p + (y cos (3 sin 6) Syp = (1 - cos &) aSap p cos 6 - Yap sin 0. Forming the quantities m, m l , m a as usual, we have < 3 - (1 + 2 cos 6} tf + (1 + 2 cos 0) < - 1 = 0, or (<-l)(< 2 -2cos0< + l) = 0, or (0 - 1) (< -cos - J^l sin 0} (< - cos 9 + J^l sin 6) = 0. Now - (< - 1) p = (1 - cos 0) (aSap + p)- sin & Fap, - (<f> - cos - J- 1 sin 6) p = (1 - cos 0} aSap + sin 6 (p A/-T- Yap), - (< - cos + J- 1 sin 6) p = (1 - cos 0) aSap - sin 6 (p J-l + 206 QUATERNIONS. [CHAP. To detect the components which are destroyed by each of these factoi's separately, we have, by (n.), for (< 1), the vector (<t>*- 2 cos e < + 1) p = - ZaSap (1 - cos ff) ; so that (< 1) a = 0, which is, of course, true. Again which we leave to the student to verify. The imaginary directions which correspond to the unreal roots are thus, in this case, parallel to the Bivectors Here, however, we reach notions which, though by no means difficult, cannot well be called elementary. A very curious case, whose special interest however is rather mathematical than physical, is presented by the assumptions for then <f>p = ({3 + y) Sap + (y + a) Sfo + (a + ft) Syp l3 + y')p- (aSap where 8 is a known unit vector. This function is obviously self- conjugate. Its cubic is < 3 - 30 + 2 = = (< - I) 2 (< + 2), which might easily have been seen from the facts that 1st, 08 = -2S, 2nd, <a = a, if SaS = 0. The case is but slightly altered when the signs of a', /3', y are changed. Then <f>p = 3S$Sp p, and the cubic is < 3 - 3<- 2 = (< + I) 2 (<- 2) = 0. X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 207 These are mere particular cases of extension parallel to the single axis 8. The general expression for such extension is obviously <}>p = p and we have for its cubic We will conclude our treatment of strains by solving the following problem : find the conditions which must be satisfied by a simple shear which is capable of reducing a given strain to a pure strain. Let < be the given strain, and let the shear be, as above, f*>l+/9&.a, then the resultant strain is Taking the conjugate and subtracting, we must have = i/<>- '' = <>-<j>' + S.<j>'a so that the requisite conditions are contained in the sole equation 2e=V<l>'a(3. This gives (1) ./3e = 0, (2) S^'a(=0 = Sa4e. But (3) Sap (by the conditions of a shear), so that xa= V . fifa. Again, (4) 2e 2 = S . <'a p f = S.a or -ma=2V.p- l <l>f. Hence we may assume any vector perpendicular to e for /?, and a is immediately determined. 208 QUATERNIONS. [CHAP When two of the roots of the cubic in < are imaginary let us suppose the three roots to be Let ft and y be such that * (ft + y J=l) = (e, + * 3 7 Then it is obvious that, by changing throughout the sign of the imaginary quantity, we have < 08 - 7 J- 1 ) = (** ~ ** JPb (ft -7 N/^l ) These two equations, when expanded, unite in giving by equating the real and imaginary parts the values To find the values of a, /?, y we must, as before, operate on any vector by two of the factors of the cubic. As an example, take the very simple case . <f>p = e Vip. Here it is easily seen by (m), (s), that in = 0, m l = + e", m a = 0,- so that < 3 + e 2 < = 0, that is As operand take then \\ || (_ jy _ fo + p ) II*. Again II - jy - Te + J - 1 (% - 11,/y + &* - ^/"^T (; - ley). X.] VECTOR EQUATIONS OF THE FIRST DEGEEE. 209 With a change of sign in the imaginary part, this will repre- sent so that /? =jy + kz, 7 =jz - ky- Thus, as the student will easily find by trial, /3 and y form with a, a rectangular system. But for all that the system of principal vectors of </>, viz. a, (3yJ^l does not satisfy the conditions of rectangularity. In fact we see by the above values of /3 and y that It may be well to call the student's attention at this point to the fact that the tensors of these imaginary vectors vanish, for This gives a simple example of the new and very curious modifications which our results undergo when we pass to Si- vectors ; or, more generally, to Biquaternions. As a pendant to the last problem we may investigate the relation of two vector-functions whose successive application produces rotation merely. Here < = ^x~ l is such that by (w) or xx = l A = since each of these functions is evidently self-conjugate. This shews that the pure parts of the strains ty and x are the same, which is the sole condition. One solution is, obviously, X ' = x -', tf = r>, T. Q. 14 210 QUATEKNTONS. [CHAP. i e. each of the two is itself a rotation ; and a new proof that any number of successive rotations can be compounded into a single one may easily be given from this. But we may also suppose either of i/r, x> suppose the latter, to be self-conjugate, so that or ur \I/ = Y , which leads to previous results. EXAMPLES TO CHAPTER X. 1. If a, /?, y be a rectangular unit system S. Va<k<iVB<l>BVy<l>y = -mS. B^'-^ai, and therefore vanishes if <f> be self-conjugate. State in words the theorem expressed by its vanishing. 2. With the same supposition find the values of SF. Fa<a. Vfi<j>{3 and of 2S . Fa<ctF/3</3. Also of 2 . aSa<j>a. 3. When are two simple shears commutative ? 4. Expand -^ - - in powers of <i. and reduce the result to 1 e</> three terms by the cubic in <. 5. Shew that *T. ^V = V ' P<PP<P P 6. Why cannot we expand <' in terms of <, <f>, </> 21 ? 7. Express Vp<f>p in terms of p, <^p, <f> 2 p, and from the result find the conditions that <p shall be parallel to p. X.] VECTOR EQUATIONS OF THE FIRST DEGREE. 211 8. Given the coefficients of the cubic in <, find those of the cubics in < 2 , </> 3 , &c. <^ n . 9. Prove 10. Ifm = .4, b, c a, B, c' a', b', C shew that M g = may be written as 11. Interpret the invariants m 1 and m a in connexion with Homogeneous Strain. 12. The cubics in tyr and \j/<f> are the same. 13. Find the unknown strains < and ^ from the equations 14. Shew that the value of F (<{>a-x a + 4>PxP + ^7X7) ^ ^ e same, whatever rectangular unit system is denoted by a, /?, y. 15. Find a system of simple shears whose successive applica- tion results in a pure strain. 16. Shew that, if < be self-conjugate, and , rj two vectors, the two following eqxiations are consequences one of the other : From either of them we obtain the equation : 142 212 QUATERNIONS. [CHAP. X. 17. Shew that in general any self-conjugate linear and vector function may be expressed in terms of two given ones, the expression involving terms of the second order. Shew also that we may write < + z = a (ps + x) 3 + b (OT + x) (w + y) + c (w + yf t where a, b, c, x, y, z are scalars, and &, w the given functions. What character of generality is necessary in ta and o> ? How is the solution affected by non-self-conjugation in one or both 1 ? 18. Solve the equations : (a) V.app=7.ayp, (b) (c) (d) apa~ l + Ppft' 1 = ypy (e) APPENDIX. WE have thought it would be acceptable to many students if we should give as an Appendix a brief, and in some cases even a detailed, solution of the most important and most difficult of the ADDITIONAL EXAMPLES. In doing so, we would add as a word of advice, that our solutions be employed simply for the purpose of comparison with those which shall occur to the student himself. CHAP. II. Ex.4. If AB = a, BC = (3, AP = ma, AP' = m'a, BQ=mf5, &c. ; then AE = AP + xPQ = AP' + x'P'Q gives ma. + x {(1 - m) a + m(3} = m'a + x' {(1 - ra') a + m'ft}, whence x = m' } and PE = m'PQ. Ex. 6. ABCD is a quadrilateral; AB = a, AC = /3, AD=y, AP = ma, BQ = m(fi- a), &c. The condition PQ + ES = gives (1 - m) a + m (ft a) + (1 ra) (y - ft) - my = 0, or (l-2ro)(a-j8 + y) = 0; an equation which is satisfied either when l-2m = 0, or when a-yS + y-0. The former solution is Ex. 5j the latter gives ABCD a parallelogram. 214 QUATERNIONS. Ex. 10. Let a, b, c be the points in which the bisectors of the exterior angles at A, B, C meet the opposite sides. Let unit vectors along BC, CA, AB be a, (3,y; then with the usual nota- tion we have (1). Now Aa = x (/8 + y) = bfi + y (bj3 + cy) be ~b-c &- , . be /0 and Aa=i (p + y). A * v / ' Similarly a-6 v be therefore Ab = (3, (by 1), c-cr v 7 Ac= y. a- b ' Hence (b c)Aa + (c - a) Ab + (a b)Ac = 0, and also ( b c) + (c a) + (a 6) = 0, therefore (Art. 13) a, b, c are in a straight line. COE. ba : ca :: b a : c-a. Ex. 12. If the figure of Ex. 11, Art. 23, be supposed to re- present a parallelepiped; then, with the notation of that example, the vector from to the middle point of OG is ^ (a + ft + 8), which is the same as the vector to the middle point of AF, viz. APPENDIX. 215 Ex. 13. "With the figure and notation of Art. 31, the former part of the enunciation is proved by the equation 1 a Also, if the edges AB, BO, CA be bisected in c, a, b, the mean point of the tetrahedron Oabc is evidently l/a + P y + a\ 2 /' 2 ~T 2 which proves the latter part of the emmciation. Ex. 14. Here we have to do with nothing but the triangles on each side of OD. IfOQ = a, QA=pa, AP = p,PD = q{3; gives pq - 1 * Similarly, if OS = a, SB=p'a!, R T'0 = x'OD 1 gives x = p'q'-V But the data are - = , p = mq; hence pq=p'q', and x = x'; therefore T coincides with T. Ex. 15. we shall have, by making A0 = AP + PO = AR + RO, therefore p + q + r = 2. Ex. 1 7. Let RA = a, RB = p, AP = ma, AD=pa + qp; then PD =pa + qP~ ma, 216 QUATERNIONS. and ES = P + PS=BQ + QS gives (1 + m) a + x (pa. + qfi- ma) = (1 + m) (3 + y (pa + q(3- m/3), 1 + m whence x , in 1 +m 1 + m . .. and JRS= (pa + qB) = - AD. m m [Or thus : QA =(l-m)a; CHAP. III. Ex. 5. Let ABGD be the quadrilateral; DA, DB, DC, a, (3, y respectively. a)+(y-a) = y (-a) + (- a)y Taking scalars, and applying 22. 3, there results, which is the proposition. Ex. 6. If a, /?, y be the vectors OA, OJB, OC corresponding to the edges a, b, c ; we have = abk + bci + caj, the negative square of which is the proposition given. APPENDIX. 217 Ex. 7. If Sa(/3-y) = Q and Sj3(a-y) = Q, then, by sub- traction, will /Sy (a /?) = 0. Ex. 8. If a 2 = ((3 - y) 2 , P* = (y- a) 2 , / = (a - /3) 2 ; then will for these are the same equations in another form; and they prove that the corresponding vectors are at right angles to one another. Ex. 9. If OA, OB, 00, OD are a, (3, y, S; triangle DAB : DAG :: tetrahedron ODA B : ODAC :: triangle OAB : OAC, because the angles which S makes with the planes OAB, OAC are equal. CHAP. IY. Ex. 1. Let be the middle point of the common perpendi- cular to the two given lines ; a, a, the vectors from to those lines, unit vectors along which are ft, y ; p the vector to a point P in a line QR which joins the given lines j P being such that RP=mPQ; therefore p + a yy = m (a + xfi p). Now since a is perpendicular to both /? and y, the equation gives (l+m) Sap = (m - 1) a 2 a plane. Ex 2. Retaining what is necessary of the notation of the last example, let OS - 8. If PR perpendicular on y meet ft in Q, we have a + yy + RP = p, which gives yy 3 = Syp ; RQ = 2 a 4- xfi yy, which gives yy* xS{3y ; 218 QUATEKNTONS. and SP a = e*PQ* gives which being of the second degree in p shews that the locus is a surface of the second order. See Chap. VI. Ex. 3. The equation of the plane is *Syp = a, which, being substituted in the equation of the surface, gives what is obviously the equation of a circle. Ex. 4. With the notation of Ex. 1, let 8, 8' be the perpen- diculars on the lines, then p+S = a + xp gives F/3S = - F/3 (p - a), and the condition given may be written Now (22. 9) whence p 2 - ZSap + a" + S*/3p = e* (p 2 + 2Sap + a 2 + S 2 y P ), a surface of the second order. Ex. 6. Sp (ft + y) = c, a plane perpendicular to the line which bisects the angle which parallels to the given lines drawn through make with one another. Ex. 7. a, (3 the vectors to the given points A, B, Syp = a, SSp = b the equations of the planes, y, 8 being unit vectors. xy, y8 the vector perpendiculars from A on the planes, then x = Say a, y Sa8 b, Sa(y + S)-(a+b) ................ (1). APPENDIX. 219 Hence by the question or S(p-a)(y + S)=Q ........................ (2). Now equation (1) will give the sum of the perpendiculars on the planes from any other point in the line AB by simply writing a + z (/3 a) in place of a ; and from equation (2) this will pro- duce no change. Ex. 8. If /3' be the vector to C, equation (2) of the last example gives Now the sum of the perpendiculars from any other point in the plane will be found from equation (1) by writing a + z (/B - a) + z' (ft' - a) in place of a. Hence the proposition. Ex. 10. Tait's Quaternions, Art. 213. Ex. 11. Let a, (3, y, 8 be the vectors OA, OB, OC, OD then (34. 5, Cor.) 8 - S. a/2y . (Fa/3 + F/?y + Fya)" 1 abc (bci + caj + abk) /, * = (a*)* +{&$)* +()* ' Now triangle ABD : triangle ABC :: tetrahedron OABD : tetrahedron OABG : : S. a/38 : S. a/3y : : S . aJrijb : S . dbcijk :: (ab)* : (ab)> + (be} 3 + (ca)* : : (triangle AOB)* : (triangle ABC) 3 . (Chap. III., Additional Ex. 6.) 220 QUATERNIONS. Ex. 12. This is merely the equation 8 p = at + ^, with t eliminated by taking the product of Vap, V(3p. (See 55. 3.) CHAP. V. Ex. 3. Let a, a' be the radii of the circles ', a, p the vectors from the centre of one of them to that of the other, and to the point whose locus is required ; then a a Ex. 7. This is the polar reciprocal of Ex. 3, Art. 40. Ex. 8. Let A be the origin, AB=B, AC = y, the vector to the centre a : then - V(AB . EC . CA) = V . (y - ft) y = y*(3-(?y = 2BSay - 2ySaB from the circle; ,-. S.a7(AB.JBO.CA) = 0. Ex. 9. Tait, Art. 222. Ex. 10. Tait, Art. 221. Ex. 11. Tait, Art. 223. Ex. 12. Tait, Art. 232. CHAP. VI. Ex. 1. Let 8 be the vector to the given point, TT the vector to the point of bisection of a chord, B a vector parallel to the chord, all measured from the centre ; then (48); APPENDIX. 221 from which, by making we get 8p<!>P 7 an ellipse whose centre is at the point of bisection of the line which joins the given point with the centre of the given ellipse. Ex. 2. Let 26 be the shortest distance between the given lines ; their angle of inclination ; 2a the line of constant length ; then as in Ex. 2, Chap. IV., the former gives a 2 + 2/ 3 -2aycos0 = 4(a'-& 2 ) .................. (1), the latter 4p = (* + y} (ft + y) + (x - y} (/? - y), which, since ft + y, ft y are vectors bisecting the angles between the lines and therefore at right angles to one another, is an equa- tion of the form of that in Art. 55. 2 ; whilst equation (1) satisfies the condition which is requisite for an ellipse. Ex. 3. Let a be a vector semi-diameter, parallel to a chord through ; 8 the vector to : then p = 8 + xa gives S<8 + 2xS8<f>a + x 2 Sa<f>a = 1, which, since >Sa<a=l, shews that the product of the two values of x is constant ; hence the rectangle by the segments of the chord varies as a*, which is the proposition. 222 QUATERNIONS. Ex. 4. With the usual notation, let CE, CE' be semi- diameters parallel to DP, D'P, and let their vectors be ra (a (3), n (a + ft) ; then since P, D, E, E' are points in the ellipse, .'. 2m 2 =1. Similarly 2n s = 1, m = n, and DP : D'P :: !>-/?) : F(a + p) :: Tm(a-(3) : Tn(a+ ft) : : CE : CE'. COR. Since m = ~, CE : DP : : 1 : J2. v j Ex. 5. Put no.', np' in place of a, p in equation (1), Art. 43. Ex. 6, 7. With everything as in Ex. 4, CE, CE' being now semi-diameters in the direction of diagonals of the parallelogram, = 0; hence CE, CE' are conjugate. Ex. 8. S (a + ft) < (a + /3) = 2 gives an ellipse, whose equation is Sp4fp = l, where <'=|; hence the diameters of the locus are to those of the given ellipse Ex. 9. If y be a unit vector to which the lines are parallel, p, p' points in which the lines cut the ellipse, and "Sp^P = 1 gives 2aSi<f>y + mSyijyy = 0| . Similarly 2bSj<}>y + nSy<j>y = 0) .................. ^ '' APPENDIX. 223 Now Sp<j>p' = an Si<f>y + bmSj(f>y + mnSy$y = 0, by equations (1) ; .*. p, p are conjugate. COR. The same demonstration applies when the diameters from whose extremities parallels are drawn, are any conjugate diameters whatever, i, j being parallel to those diameters. Ex. 10. Let CP, CP 1 be any two semi-diameters, their vec- tors being a, a'; PQ the semi-ordinate to CP'; CQ na! ; then S (PQ . <j>a?) = gives S (a - na) <f>a? = 0, .*. n = /Satf>a. Now the area of the triangle QGP is proportional to V(CP.CQ), i. e. to n Vaa or to Sa<f>a . Faa', which, being symmetrical in a, a', proves the proposition. Ex. 11. If the tangent at P' meet CP produced in F, CT=ma; then, since PT is perpendicular to <j>af, -^r>> oa<pa and area P f CT is proportional to V(CP f .CT), i.e. to which is symmetrical in a, a'. Ex. 12. Let a, ft be the vector semi-diameters of the larger ellipse ; G the centre ; the centre of the smaller ellipse, whose equation is = c 224 QUATERNIONS. y a vector along PQR ; then c _ - 2 ' and since CQ = a + (3+ xy, hence PR is conjugate to CQ, and therefore bisected at Q. Ex. 13. This is simply a combination of 49. 2 and 49. 1. CHAP. VII. Ex. 3. The equation of the circle is 9 which by 52. 1 gives (a? Sap) 2 a'Sap = ^ a , ID 2 . '. Sap = r , 4 which (52. 11) is the proposition. Ex. 5. If be the centre of the circle, Q a point at which it meets the tangent at A ; then, with the notation of 55. 1, i. e. z* zy + ^- = 0, which gives two equal values of z ; hence the proposition. APPENDIX. 225 Ex. 6. With any point as origin, let (3, y be the vectors to the two given points, TT the vector to the focus of one of the parabolas. Write aa in place of a in equation (1), Art. 52, a being a unit vector ; then - (p -)' = {a + Sa(p -v)}' .................... (1) whence, by subtraction, /3 2 - y 2 - 2S-H- (ft - y) = - Sa (J3 - y) {2a + Sa (ft - y) - 2Sair}, which gives a by a simple equation in TT; and then equation (1) becomes a quadratic in TT. Ex. 8. If two tangents meet at T, it is easy, as in Ex. 5, Art. 55, with the notation available for the focus, to find , a =- a H -- ^ p aa, 4a 2 and S(ST. ST') = will follow at once, from the fact that Ex. 9. Let P be the point of contact, PQ the chord, TEF the line parallel to the axis cutting the curve in Ej ; E the origin ; .,jr = 2 a -i- *p, J^J. = - ^ , F P Fl i pz> -K 1 -*' 2 '\o) . j JL J> / 2/ a -^ 2 / J < tt' whence- z = - f r/ > y = - , t t .-. PF : FQ :: t : t' ? tt?_ :: 2 : "2 :: TE : EF. Ex. 10. This is evident from equation (1), Art. 52. T. Q. 15 226 QUATERNIONS. Ex. 11. With the notation of Art. 52, let . '. x (a. 2p) = a + 8, x (a 2 - 2ap) - a 2 . But p, xp being vectors to the parabola, equation (1), Art. 52, gives . '. X (a 2 Sap) = a 2 + OS/Sap, X (a 2 - 2Sap) = a 2 , .'. x=x', and the proposition is true (Euc. VI. 2). Ex. 14. Tait, Art. 43, Cor. 2. Ex. 15. CP= at + - gives CF= 2at, t so that the equation of RQPSf is whence for B and R' the values of x are 2 and 1 ; therefore C=3at, Ctf^l^, 2> t QR=at-^ = PQ = APPENDIX. 227 Ex. 1 6. If CR = aa ; a + m(3, a. - m(3 vectors parallel to the given conjugate diameters, CP = aa + x ( CD = aa. + SB' (a mjS) =at' ^ , tr give t = t'; therefore CP, CD are conjugate. Ex. 18. Adopting the figure and notation of Ex! 2 of the hyperbola, Art. 55, we have t therefore R - T) (to. - *-\ , and rQ. QR= (X s - Y 3 ) (to. - |Y = PO a , since ^ 8 -F 2 =l. As an example of combining not merely the forms but the results of the Cartesian Geometry with Quaternions, we will add one more example. If CP, CD; CP, CD 1 be two pairs of conjugate semi-diameters of an ellipse, PD' will be parallel to P'D. Let CP, CP be denoted, as in Art. 55. 2, by xa + yp, x'a + y r p respectively; then CD, CD' will be represented by b _ a , b ln with the conditions aY + 6V = a*b a , a*y" + b*x' 2 = a*b* ............ (1 ). Now vector D'P = (x + 1 y' j a 4- ( y - - x J /?, 152 228 QUATERNIONS. But equations (1) g r .ve, by subtraction, a . b , a , b x + 7 y : V x :: x + -rV ' V x b j a b y a therefore D'P is a multiple of DP' and consequently parallel to it. COR. PD' : P'D : : ay' + bx : ay + bx. CHAP. VIII. Ex. 1. With the notation of Additional Ex. 1, Chap. IV., the perpendiculars are p - a - xp, p + a - yy, so that Sfip = xft 2 , Syp = yy 2 ; and by the question, (p - a - p-^SppY = e* (p + a - y^Syp)*, a surface of the second order in p. Ex. 3. The equations Sp<f>p= 1, Sir<j>p = 1, with the condition Tr = X(f>p, give 1 7T 2 j STT(J>~ l ir = 1 , = 1 respectively, X X therefore Sir^T 1 -* = IT*, whence the Cartesian equation. Ex. 4. If a, p, y are the vector radii, Sa<f>a(SiU'a) 2 (SjUa) 2 (Ta}*~ a 3 &c. = &c, Adding and observing that Sa<f>a- 1, &c., there results 1 1111 = ~i + Ii + -2 ' a o c APPENDIX. 229 Ex. 5. As in Ex. 8, Art. 64, <s? <*> and if vector OQ l = xfa, the ellipsoid gives Now ri = O*.OQ* x* and, since (Ex. 7, Art. 64), the result required is obtained by simply adding. Ex. 6. Let pk be the vector distance from the origin, of the plane parallel to xy, IT a point in it; then Sk (TT pk) = gives 8-rrk = const. Now Spffrir = 1 is the equation of the plane of contact, and if zk be the point in which this plane cuts the axis of z, zSk<j>ir = 1, i.e. zSir<j>k = 1, gives z. Now tj>k is a multiple of k, and since Sirk is constant, z is constant. Ex. 7. The equations of the ellipsoids Sp<f>p = 1, S (p - a) < (p - a) = 1, give Sp(f>a = const, as the plane of contact. Ex. 8. If pa be the vector to the point in the line OA ; the equation of its polar plane is Spa<f>p = 1 ; and the square of the reciprocal of the perpendicular from the centre on this plane is 2 . Hence the conclusion by Ex. 8, Art. 64. Ex. 9. Let p be the vector to P ; a, (3, y vector radii parallel to the chords ; then p + xa, p + y(3, p + zy, 230 QUATERNIONS. will be the vectors to A, , C ; and since P, A, JB, C are points in the ellipsoid 0, 2/S/Dj3 + y = 0, + 2 = 0. The equation of the plane ABC is (34. 5) S . (TT p) (xya-P + yzfiy + zxya) = xyzS . a/2y, and since a, /?, y are at right angles to one another, therefore the equation of the plane ABC becomes <7 ( which is satisfied by TT p = m(f>p, where and therefore Ex. 4 above gives 2 CHAP. IX. Ex. 2 and 3. Employ formula 11. Ex. 5. Since formulae 4 and 6 give the required result. Ex. 6. Apply formula 10 to Ex. 5. Ex. 8. (a/3y) 2 = a/3y . a/3y = afiy (S.afty + V. a/?y) APPENDIX. 231 Ex. 9. Formula 10 gives the vector of the product of three vectors a, /?, y, under the form a' /3' + y' where a = aSpy, &c. Hence the required scalar may be written and as the scalar part of this product is that which involves all of the three vectors a', /?', y we have exactly as in the demonstra- tion of formula 5, -a. a - ', P, -y -', p 1 , y 10. The scalar part, by formula 16, is reduced to SaSSpy - SaySpS - SaSSpy + SafiSyS + SaySfty - SapSyS, which is identically 0. The vector part, by formula 12, is aS. y&p-pS.yfa + aS. 8(3y-yS. Sfia + cuS. (3y8-SS. (3ya, which, by formula 13, reduces to 2aS. 12. If, for brevity, we denote S. a(3y, V . a(3y respectively by S and V, we have, by formula 7, 2aj8Y + a 2 (Py)* + ft* (ay) 2 + y 2 (a/3)' - (a/3y) 2 = 2aj8y . yj8a + (3ya . a/3y + ay@ . flay + a/?y . ya.fi - (afty) 2 Y)(S-V+2ySap) - (S+ V) 3 The student is recommended to verify a few examples such as the above; by putting a = i, P = ai + bj + ck, y = a'i + b'j + c'k, 232 QUATERNIONS. with the conditions a* + b 3 + c a = l, a' 2 + 6' 2 + c' 2 = l. The quaternion equality will then reduce itself to four alge- braic equalities, one of which is obvious, and the others are p* + r a a' 2 a 2 + 2aa'm = 0, pq - mr + a'c' + ac- lac'm = 0, qr + mp + a'b' + ab 2ab'm - 0, where m = aa' + bb' + cc', p = ab' - a'b', q = be' b'c, r = ca c'a. Ex. 13. S. (a -8) (ft- 8) (y - 8) = S. aj8y - S . fa8 + S. yBa - S. Sap. Ex. 14. By 34. 8, we have aS.83 BCD therefore the same Article ives and since the scalar of the product of this vector by the vector perpendicular to the plane in which A, B, C, D lie gives the right- hand side of Ex. 13, we obtain a . BCD - ft . CDA + y . DAB - 8 . ABC = 0. CAMBRIDGE : PRINTED BY C. J. CLAY, M.A. AT THE UNIVERSITY PRESS. April 1892 A Catalogue OF Educational Books PUBLISHED BY Macmillan & Co. BEDFORD STREET, STRAND, LONDON For books of a less educational character on the subjects named below, see Macmillan and Co.' s Classified Catalogue of Books in General Literature. CONTENTS GREEK AND LATIN CLASSICS- ELEMENTARY CLASSICS . CLASSICAL SERIES . CLASSICAL LIBRARY ; Texts, Com mentaries, Translations GRAMMAR, COMPOSITION, AND PHI LOLOGY ANTIQUITIES, ANCIENT HISTORY, AND PHILOSOPHY MODERN LANGUAGES AND LITERATURE- ENGLISH FRENCH GERMAN MODERN GREEK . 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